# NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium- In this chapter, you will get the NCERT solutions for class 11 chemistry chapter 7 Equilibrium of chemical and physical processes and details about equilibrium's dynamic nature. There are some insights about the equilibrium constant, the law of mass action and what are the factors affecting the equilibrium state. It is a lengthy and also an important chapter of class 11 chemistry. In NCERT solutions for class 11 chemistry chapter 7 Equilibrium, there are 73 questions in the exercise. The solutions of NCERT class 11 chemistry chapter 7 Equilibrium are prepared and solved by chemistry experts. These NCERT solutions will help you in the preparation of your class 11 final examination as well as in the various competitive exams like JEE Mains, NEET, BITSAT, etc.

After studying NCERT solutions for class 11 chemistry chapter 7 Equilibrium, you will be able to identify equilibrium's dynamic nature involved in chemical and physical processes; explain the law of equilibrium; explain characteristics of equilibria involved in chemical and physical processes; write expressions for equilibrium constants; explain factors affecting the equilibrium state of a reaction; classify substances as bases or acids according to Bronsted-Lowry, Arrhenius and Lewis concepts; also able to classify bases and acids as weak or strong in terms of their ionization constants and calculate solubility product constant.

Exercise Questions

## What is the Equilibrium?

Equilibrium is the state of a process in which the properties like the concentration of the system, pressure, and temperature do not change with the passage of time. It can be established for both chemical and physical processes. At the stage of equilibrium, the rate of forwarding and reverse reactions are equal. The state of equilibrium can only be achieved if the reversible reaction is taking place in a closed system.

 Important formulas of NCERT class 11 chemistry chapter 7 Equilibrium    1.Equilibrium constant, K              $K=\frac{K_f}{K_b}$ 2. Concentration quotient, Q          $Q=\frac{[C]^c[D]^d}{[A]^a[B]^b}$ 3. $\Delta G^o=-2.303RT\:logK$ 4. $K_w=[H^+][OH^-]$ 5. $pH = -log[H^+]$

## Question 7.1(a)        A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

What is the initial effect of the change on vapour pressure?

By increasing the volume of the container suddenly, initially, the vapour pressure would decrease. It is due to, the amount of vapour has remained the same but it is distributed in a larger volume.

How do rates of evaporation and condensation change initially?

Here the temperature is constant so that the rate of evaporation is also the same as before. On increasing the volume of the container, the density of vapour decreases due to which the rate of collision between vapour particles decreases. Hence the condensation rate also decreases initially.

What happens when equilibrium is restored finally and what will be the final vapour pressure?

When equilibrium is restored, the rate of evaporation is equal to the rate of condensation. The temperature is constant and the volume is changed. The vapour pressure is temperature-dependent, not volumes. So, that final vapour pressure is equal to the initial vapour pressure.

$2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)}$

Followings are the given information values to solve the above problems-
$\\{[SO_{2}]}= 0.60 M\\ {[O_{2}]}=0.82M\\ {[SO_{3}]}=1.90M$
The given chemical equation is -

$2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)}$

The equilibrium constant for this reaction is expressed as;
$K_c=\frac{[SO_{3}]^2}{[SO_{2}]^2[O_{2}]}$

$\\=\frac{[1.90]^2}{[0.60]^2[0.82]}\\ =12.22 M^{-1}$(approx)

$I_{(g)}\rightleftharpoons 2I_{(g)}$

Calculate Kp for the equilibrium.

It is given that total pressure (PT)is $10^5\ Pa$ and partial pressure of $I$ atom = 40 % of PT
So, the partial pressure of $I$ atom = $\frac{40}{100}\times 10^5 = 4\times 10^4$Pascal

The partial pressure of $I_2$ = 60% of PT

So, the partial pressure of $I_2$ = $\frac{60}{100}\times 10^5=6\times 10^4$Pascal

Now, for the reaction  $I_{(g)}\rightleftharpoons 2I_{(g)}$
$K_p = \frac{(p_I)^2}{p_{I_2}} = \frac{(4\times 10^4)^2}{6\times 10^4}\ Pa$
$=2.67\times 10^4\ Pa$

(i)      $2NOCl_{(g)}\rightleftharpoons 2NO_{(g)}+Cl_{2}_{(g)}$
(ii)    $2Cu(NO_{3})_{2}_{(s)}\rightleftharpoons 2CuO_{(s)}+4NO_{2}_{(g)}+O_{2}_{(g)}$
(iii)     $CH_{3}COOC_{2}H_{5}_{(aq)}+H_{2}O_{(l)}\rightleftharpoons CH_{3}COOH_{(aq)}+C_{2}H_{5}OH_{(aq)}$
(iv)     $Fe^{3+}_{(aq)}+3OH^{-}_{(aq)}\rightleftharpoons Fe(OH)_{3}_{(s)}$
(v)       $I_{2}_{(s)}+5F_{2}\rightleftharpoons 2IF_{5}$

The equilibrium constant for any reaction can be written as (concentration of products) / (concentration of reactants). And we considered constant values for the solids and liquids because their density per unit volume or mass per unit volume does not change.
Thus,
(i)
$K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$
(ii)
$K_c = \frac{[CuO]^2[NO_2]^4[O_2]}{[Cu(NO_3)_2]^2}$
(iii)

$K_c = \frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H5]}$

(iv)
$K_c = \frac{1}{[Fe^{3+}][OH^-]}$

(v)
$K_c = \frac{[IF_5]^2}{[F_2]^5}$

(i)     $2NOCl _{(g)}\rightleftharpoons 2NO_{(g)}+Cl_{2}_{(g)};K_{p}=1.8 \times 10^{-2} at 500K$

(ii)     $CaCO_{3}_{(s)}\rightleftharpoons CaO_{(s)}+CO_{2}_{(g)};K_{p}=167 at 1073K$

We know that the relation between $K_p$ and $K_c$ is expressed as;

$K_p=K_c(RT)^{\Delta n}$............................(i)
here ${\Delta n}$ = (no. of moles of product) - (no. of moles of reactants)

R = 0.0831 bar L /mol/K, and

For (i)
$K_p$ = $1.8\times 10^{-2}$ and Temp (T) = 500K

${\Delta n}$ = 3 - 2 = 1
By putting the all values in eq (i) we get

$K_c = \frac{1.8\times 10^{-2}}{0.0831\times 500} = 4.33\times 10^{-4}$

For (ii)
$K_p$ = $167$ and temp(T) = 1073 K
${\Delta n}$ = 2 - 1 = 1
Now, by putting all values in eq (i) we get,

$K_c = \frac{167}{0.0831\times 1073} = 1.87$

Question

$NO_{(g)}+O_{3}_{(g)}\rightleftharpoons NO_{2}_{(g)}+O_{2}_{(g)}$

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

It is given that,
$K_c=6.3\times 10^{14}$
we know that $K'_c$ for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:

$K'_c=\frac{1}{K_c}$
$=\frac{1}{6.3\times 10^{14}}$
$=1.58\times 10^{-15}$

For the pure liquids and solids, the molecular mass and the density at a particular temperature is always fixed and it is considered as a constant. Thus they can be ignored while writing the equilibrium constant expression

Question

$2N_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2N_{2}O_{(g)}$

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 ×10-37, determine the composition of equilibrium mixture.

It is given that,
$K_c = 2.0\times 10^{-37}$
Let the concentration of $N_2O$ at equilibrium be $x$. So,

$2N_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2N_{2}O_{(g)}$
initial conc                  0.482        0.933                     0          (in moles)
at equilibrium             0.482-$x$   0.933 -  $x$              $x$           (in moles)

The equilibrium constant is very small. So, we can assume   0.482-$x$ = 0.482 and 0.933 -  $x$ = 0.933

We know that,

$K_c = \frac{[N_2O]^2}{[N_2]^2[O_2]}$
$2\times 10^{-37}=\frac{(x/10)^2}{(0.0482)^2(0.0933)}$       {dividing the moles by 10 to get concentration of ions)

$\\\frac{x^2}{100}= 2\times 10^{-37}\times (0.0482)^2\times (0.0933)\\ x^2 = 43.35\times 10^{-40}\\ x = \sqrt{43.35\times 10^{-40}}\\ x=6.6\times 10^{-20}$

So, the concentration of $[N_2O]= \frac{x}{10}=6.6\times 10^{-20}$

$2NO_{(g)}+Br_{2}_{(g)}\rightleftharpoons 2NOBr_{(g)}$

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br2  .

The initial concentration of $NO$ and $Br_2$  is  0.087 mol and 0.0437 mol resp.

The given chemical reaction is-
$2NO_{(g)}+Br_{2}_{(g)}\rightleftharpoons 2NOBr_{(g)}$

Here, 2 mol of $NOBr$ produces from 2 mol of $NO$. So, 0.0518 mol of $NOBr$  is obtained from 0.518 mol of $NO$.

Again, From 1 mol of $Br_2$  two mol of $NOBr$ produced. So, to produce 0.518 mol of $NOBr$ we need $\frac{0.518}{2} = 0.0259$ mol of $Br_2$.

Thus, the amount of $NO$ present at equilibrium = 0.087 - 0.0518 = 0.0352 mol

and the amount of $Br_2$ present at the equilibrium = 0.0437-0.0259 = 0.0178 mol

Question

$2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)}$

What is Kc at this temperature?

We have,
$K_p= 2\times 10^{10}/bar$
We know that the relation between $K_p$ and $K_c$;

$K_p=K_c(RT)^{\Delta n}$
Here $\Delta n$ = ( moles of product) - (moles of  reactants)

$2SO_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2SO_{3}_{(g)}$

So. here $\Delta n$ = 2-3 = -1

By applying the formula we get;

$2\times 10^{10} =K_c(0.0831 L\ bar/K/mol)\times 450K$
$K_c= \frac{2\times 10^{10}}{(0.0831 L\ bar/K/mol)\times 450K}$

$=74.79 \times 10^{10}L mol^{-1}$
=  7.48$\times 10^{10}L mol^{-1}$

$2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}$

The initial pressure of $HI$ is 0.2 atm . At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of $HI$  is 0.2 - 0.04 = 0.16.

The given reaction is:

$2HI_{(g)}\rightleftharpoons H_{2}_{(g)}+I_{2}_{(g)}$

$HI_{(g)}\rightleftharpoons 1/2H_{2}_{(g)}+1/2I_{2}_{(g)}$

At equilibrium,
$\\pHI = 0.04\\ pH_2 = \frac{0.16}{2}=0.08\\ pI_2 = \frac{0.16}{2}=0.08$

Therefore,
$K_p = \frac{pH_2\times pI_2}{p^2HI}$
$\\=\frac{0.08 \times 0.08}{(0.04)^2}\\ =4$

We have,
$N_{2}_{(g)}+3H_{2}_{(g)}\rightleftharpoons 2NH_{3}_{(g)}$

$K_c = 1.7 \times 10^{2}$

The concentration of species are-
$[N_2] = 1.57/20 mol L^{-1}\\ {[H_2]}=1.92/20mol L^{-1}\\ {[NH_3]}=8.13/20mol L^{-1}$

We know the formula of

$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$
$\\=\frac{(8.13/20)^2}{(1.57/20)(1.92/20)^3}\\ =2.4\times 10^{-3}$
The reaction is not in equilibrium. Since $Qc>K_c$, the equilibrium proceeds in reverse direction.

$H_{2}O_{(g)}+CO_{(g)}\rightleftharpoons H_{2}_{(g)}+CO_{2}_{(g)}$

Calculate the equilibrium constant for the reaction.

The given reaction is-

$H_{2}O_{(g)}+CO_{(g)}\rightleftharpoons H_{2}_{(g)}+CO_{2}_{(g)}$
initial conc                   1/10            1/10              0               0
At equilibrium             0.6/10           0.6/10          0.04          0.04

Now, the equilibrium constant for the reaction can be calculated as;

$K_c = \frac{[H_2][CO_2]}{[H_{2}O][CO]}$
$= \frac{.04 \times.04}{(.06)^2}$

= 0.44 (approx)

Question

$H_{2}_{(g)}+I_{2}_{(g)}\rightleftharpoons 2HI_{(g)}$

is 54.8. If 0.5 mol L-1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

We have,

The equilibrium constant of the reaction = $54.8$

moles of $HI$ = 0.5 mol/L

The given reaction is-

$H_{2}_{(g)}+I_{2}_{(g)}\rightleftharpoons 2HI_{(g)}$

So, the reverse equilibrium constant is $K'_c = 1/K_c$

Suppose the concentration of hydrogen and iodine at equilibrium be $x$

$[I_2]=[H_2]=x$

Therefore,

$K'_c = \frac{[H_2][I_2]}{[HI]} = \frac{x^2}{(.5)^2}$
So, the value of $x$ = $\sqrt{\frac{0.25}{54.8}}=0.0675(approx)\ mol/ L$

$2ICl _{(g)}\rightleftharpoons I_{2}_{(g)}+Cl_{2}_{(g)}; K_{c}=0.14$

The given reaction is:
$2ICl (g) \rightleftharpoons I_2 (g) + Cl_2 (g)$
Initial conc.                0.78 M           0           0
At equilibrium         (0.78 - 2$x$) M     $x$ M        $x$ M

The value of $K_c = 0.14$

Now we can write,
$K_c = \frac{[I_2][Cl_2]}{[ICl]^2}$
$\\0.14 = \frac{x^2}{(.78-x)^2}\\ 0.374 = \frac{x}{(.78-x)}$

By solving this we can get the value of $x$ = 0.167

$C_{2}H_{6}_{(g)}\rightleftharpoons C_{2}H_{4}_{(g)}+H_{2}_{(g)}$

Suppose the pressure exerted by the hydrogen and ethene gas be $p$ at equilibrium.

the given reaction is-

$C_{2}H_{6}_{(g)}\rightleftharpoons C_{2}H_{4}_{(g)}+H_{2}_{(g)}$
initial pressure                  4 atm              0                     0
At equilibrium                  4 - $p$                 $p$                    $p$

Now,
$K_p = p_{C_2H_4}\times p_{H_2}/p_{C_2H_6}$
$0.04 = \frac{p^2}{4-p}$

By solving the quadratic equation we can get the value of  $p$ = 0.38

Hence,at equilibrium,

$p_{C_2H_6}$ =  4  - p  = 4 -.038

= 3.62 atm

$CH_{3}COOH _{(l)}+C_{2}H_{5}OH\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

The given reaction is-
$CH_{3}COOH _{(l)}+C_{2}H_{5}OH\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$
the concentration ratio (reaction quotient) of the given chemical reaction is-

$Q_c = \frac{[CH_{3}COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$

$CH_{3}COOH _{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$

(ii)  At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Let the volume of the mixture will be V.
$CH_{3}COOH _{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$
initial conc.                                     1/V                             0.18/V                          0                                   0

At equilibrium                          $\frac{1-.171}{V}$  (= 0.829/V)  $\frac{1-.171}{V}$               $0.171$                         $0.171$

So the equilibrium constant for the reaction can be calculated as;

$K_c = \frac{(0.171)^2}{(0.829)(0.009)}$

$=3.92$ (approx)

$CH_{3}COOH_{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Let the volume of the mixture will be V.
$CH_{3}COOH _{(l)}+C_{2}H_{5}OH_{(l)}\rightleftharpoons CH_{3}COOC_{2}H_{5}_{(l)}+H_{2}O_{(l)}$
initial conc.                                     1/V                             0.5/V                          0                                   0

At equilibrium                          $\frac{1-.214}{V}$  (= 0.786/V)  $\frac{1-.214}{V}$               $0.214$                         $0.214$

Therefore the reaction quotient of the reaction-

$Q_c = \frac{(0.214)^2}{(0.786)(0.286)}$

$=0.2037 (approx)$

Since $Q_c equilibrium has not been reached.

$PCl_{5}_{(g)}\rightleftharpoons PCl_{3}_{(g)}+Cl_{2}_{(g)}$

We have,

concentration of $PCl_5$ = 0.05 mol/L
and $K_c = 8.3\times 10^{-3}$

Suppose the concentrations of both $PCl_3$ and $Cl_2$ at equilibrium be $x$ mol/L. The given reaction is:

$PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2(g)$

at equilibrium       0.05                    $x$                     $x$

it is given that the value of the equilibrium constant, $K_c = 8.3\times 10^{-3}$

Now we can write the expression for equilibrium as:

$K_c = \frac{x^2}{0.05}=8.3\times 10^{-3}$
$\Rightarrow x = \sqrt{4.15\times 10^{-4}}$
$=0.0204$ (approx)

Hence the concentration of $PCl_3$ and $Cl_2$  is 0.0204 mol / L

We have,

$K_p=0.265$
the initial pressure of $CO$ and $CO_2$ are 1.4 atm and 0.80atm resp.

The given reaction is-

$FeO_{(s)}+CO_{(g)}\rightleftharpoons Fe_{(s)}+CO_{2}_{(g)}$
initially,                                   1.4 atm                    0.80 atm

$Q_p = \frac{pCO_2}{pCO}=\frac{0.80}{1.4} = 0.571$
Since $Q_p>K_c$ the reaction will proceed in the backward direction to attain equilibrium. The partial pressure of $CO_2$ will increase = decrease in the partial pressure of $CO_2$   = $p$

$\therefore K_p = \frac{pCO_2}{pCO} = \frac{0.80-p}{1.4+p}=0.265$
$=0.371+.265p=0.80-p$

By solving the above equation we get the value of $p$ = 0.339 atm

Question

$N_{2}_{(g)}+3H_{2}_{(g)}\rightleftharpoons 2NH_{3}_{(g)}$

At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L-1 N2, 2.0 mol L-1 H2 and 0.5 mol L-1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

The given reaction is:

$N_{2}_{(g)}+3H_{2}_{(g)}\rightleftharpoons 2NH_{3}_{(g)}$

at a particular time:           3.0molL-1  2.0 molL-1     0.5molL-1

Now, we know that,

$Qc = [NH_3]2^ / [N_2][H_2]^3$
$\\ = (0.5)^2 / (3.0)(2.0)^3\\ = 0.0104$

It is given that Kc = 0.061

Since, Qc $\neq$  Kc, the reaction mixture is not at equilibrium.

Again, $Q_c < K_c$, the reaction will proceed in the forward direction to attain the equilibrium.

$2BrCl_{(g)}\rightleftharpoons Br_{2}_{(g)}+Cl_{2}_{(g)}$

For which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10-3 mol L-1, what is its molar concentration in the mixture at equilibrium?

Suppose the $x$ amount of bromine and chlorine formed at equilibrium. The given reaction is:

$2BrCl_{(g)}\rightleftharpoons Br_{2}_{(g)}+Cl_{2}_{(g)}$

Initial Conc.     $3.3\times10^{-3}$          0            0

at equilibrium   $3.3\times10^{-3}$- 2$x$    $x$          $x$

Now, we can write,

$K_c =\frac{[BR_2][Cl_2]}{[BrCl_2]^2}$
$32=\frac{x^2}{(3.3\times 10^{-3}-2x)^2}$
$5.66=\frac{x}{(3.3\times 10^{-3}-2x)}$
$x+11.32x=18.678 \times 10^{-3}$

By solving the above equation we get,

$x$ = $1.51 \times 10^{-3}$

Hence, at equilibrium $[BrCl_2] = 3.3\times 10^{-3} - (2\times 1.51\times 10^{-3})$
$= 0.3\times 10^{-3}$ M

$C_{(s)}+CO_{2}_{(g)}\rightleftharpoons 2CO_{(g)}$

Suppose the total mass of the gaseous mixture is 100 g.
Total pressure is 1 atm

Mass of $CO$ = 90.55 g
And, mass of $CO_2$ = (100 - 90.55) = 9.45 g

Now, number of moles of  $CO$= 90.55/28 = 3.234 mol  (mol. wt of $CO$ = 28)
Number of moles of $CO_2$ = 9.45/44 = 0.215 mol  (mol. wt of $CO_2$ = 44)

Partial pressure of $CO$ ,

$pCO=\frac{n_{CO}}{n_{CO}+n_{CO_2}}p_T$

$=\frac{3.234}{3.234+0.215} \times 1$
= 0.938 atm

Similarly partial pressure of $CO_2$,

$pCO_2=\frac{n_{CO_2}}{n_{CO}+n_{CO_2}}p_T$
= 0.062 atm

Thus, $K_p = \frac{(0.938)^2}{(0.062)} \approx 14.19$

By using the relation$K_p = K_c(RT)^{\Delta n}$
$K_c = \frac{14.19}{(0.083\times 1127)^1}$
= 0.159 (approx)

$NO{(g)}+\frac{1}{2}O_{2}(g) \rightleftharpoons NO_{2}(g)$

where

$\Delta _{f}G^{+}\left [ NO_{2} \right ]=52.0kJ/mol$

$\Delta _{f}G^{+}\left [ NO \right ]=87.0kJ/mol$

$\Delta _{f}G^{+}\left [ O_{2} \right ]=0kJ/mol$

Given data,

$\Delta _{f}G^{+}\left [ NO_{2} \right ]=52.0kJ/mol$
$\Delta _{f}G^{+}\left [ NO \right ]=87.0kJ/mol$

$\Delta _{f}G^{+}\left [ O_{2} \right ]=0kJ/mol$

given chemical reaction-
$NO{(g)}+\frac{1}{2}O_{2}(g) \rightleftharpoons NO_{2}(g)$

for the reaction,
$\Delta G^0$ = $\Delta G^0$(products) - $\Delta G^0$ (reactants)
= (52-87-0)
= -35 kJ/mol

(a)        $PCl_{5}_{(g)}\rightleftharpoons PCl_{3}_{(g)}+Cl_{2}_{(g)}$

(b)     $CaO(s)+CO_{2}(s)\rightleftharpoons CaCO_{3}(s)$

(c)     $3Fe(s)+4H_{2}O(g)\rightleftharpoons Fe_{3}O_{4}(s)+H_{2}(g)$

According to Le Chatellier's principle, if the pressure is decreased, then the equilibrium will shift in the direction in which more number of moles of gases is present.

So,

• (a)The number of moles of reaction products will increase.

• (b)The number of moles of reaction products will decrease

• (c)The number of moles of reaction products remains the same.

(i)       $COCl_{2}(g)\rightleftharpoons CO(g)+Cl_{2}(g)$

(ii)    $CH_{4}(g)+2S_{2}(g)\rightleftharpoons CS_{2}(g)+2H_{2}S(g)$

(iii)      $CO_{2}(g)+C(s)\rightleftharpoons 2CO(g)$

(iv)     $2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$

(v)     $CaCO_{3}(s)\rightleftharpoons CaO(s)+CO_{2(g)$

(vi)     $4NH_{3}(g)+5O_{2}(g)\rightleftharpoons 4NO(g)+6H_{2}O(g)$

According to Le Chatellier's principle, if the pressure is increased, then the equilibrium will shift in the direction in which less number of moles of gases is present. So, as per this rule following given reactions are affected by the increasing pressure-

The reaction (i), (iii), and (vi)- all proceeds in the backward direction

Reaction(iv) will shift in the forward direction because the number of moles of gaseous reactants is more than that of products.

$H_{2}(g)+Br_{2}(g)\rightleftharpoons 2HBr(g)$

Given that,
$K_p$ for the reaction = $1.6\times 10^{5}$

$K'_p = 1/K_p = \frac{1}{1.6\times 10^{5}} = 6.25 \times 10^{-6}$

Let the pressure of both $H_2$ and $Br_2$  at equilibrium be $p$.

$2HBr\rightleftharpoons H_2(g)+Br_2(g)$
initial conc.           10                   0              0
at eq                     10-2p             p                p

Now,
$K_p' = \frac{p^{Br_2}.p^{H_2}}{p^2_{HBr}}$
$6.25\times 10^{-6}=\frac{p^2}{(10-2p)^2}$
$5\times 10^{-3}=\frac{p}{(5-p)}$
By solving the above equation we get,

$p$ = 0.00248 bar

Hence the pressure of  $H_2$ and $Br_2$  is 0.00248 bar and pressure of $HBr$ is 0.00496 bar

$CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)$

Write an expression for Kp for the above reaction.

$CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)$
the expression of ionisation constant ( Kp) for the reaction can be defined as the ratio of the product of concentration to the product of reactants.
$K_p = \frac{p_{CO}.p^3_{H_2}}{p_{CH_4}.p_{H_2O}}$

$CH_{4}(g)+H_{2}O(g)\rightleftharpoons CO(g)+3H_{2}(g)$

How will the values of Kp and composition of equilibrium mixture be affected by

(i)      increasing the pressure

(ii)      increasing the temperature

(iii)      using a catalyst?

(i) According to Le Chatellier's principle, if pressure is increased, then the reaction will shift towards the less number of moles of gases. So, here the direction of equilibrium is backward and the value of $K_p$ decreases.

(ii) According to  Le Chatellier's principle, as the reaction is endothermic, the equilibrium will shift in the forward direction. The value of $K_p$ is increases.

(iii) The equilibrium of the reaction is not affected by the presence of the catalyst. It only increases the rate of reaction.

Question 7.29(a)_        Describe the effect of :

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$
(a)According to Le Chatelliers principle, on the addition of dihydrogen, the number of mole of $H_2$ increases on the reactant side. Thus to attain the equilibrium again the reaction will move in the forward direction.

(b) According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

(c) If we remove the $CO$ from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

(d) On removal of $CH_{3}OH$ the equilibrium will shift in the forward direction.

Question 7.29(b)     Describe the effect of :

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$
According to Le Chatellier's principle, on the addition of methyl alcohol, the number of moles of methyl alcohol increases on the product sides. So, to attain the equilibrium, the reaction will proceed in a backward direction.

Question 7.29(c)       Describe the effect of :

removal of CO

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$

$2H_{2}(g)+CO(g)\rightleftharpoons CH_{3}OH(g)$
If we remove the $CO$ from the reactant side, the concentration on the reactant side will decrease and to attain an equilibrium, the reaction will shift backward direction

$PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g), \Delta _{r}H^{+}=124.0kJ mol^{-1}$

(a)       write an expression for Kc for the reaction.
(b)     what is the value of Kc for the reverse reaction at the same temperature ?
(c)      what would be the effect on Kc if

(ii)   pressure is increased
(iii)  the temperature is increased ?

We have,
$PCl_{5}(g)\rightleftharpoons PCl_{3}(g)+Cl_{2}(g), \Delta _{r}H^{+}=124.0kJ mol^{-1}$
Equilibrium constant for the above reaction = $8.3\times 10^{-3}$

(a) Expression of $K_c$ for this reaction-
$K_c=\frac{[PCl_3][Cl_2]}{PCl_5}$

(b) The value of reverse equilibrium constant can be calculated as;

$K'_c = \frac{1}{K_c}$
$=\frac{1}{8.3\times 10^{-3}}=1.20\times 10^2$

(c).i   $K_c$ would remain the same because the temperature is constant in this case.

(c). ii  If we increase the pressure,  there is no change in $K_c$ because the temperature is constant in this case also.

(c). iii  In an endothermic reaction, the value of   $K_c$ increases with increase in temperature.

$CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}+H_{2}(g)$

If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that $p_{CO}=p_{H_{2}O}$ = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

We have,
The partial pressure of $CO$ and $H_2O$ is 4 bar and the $K_p = 10.1$
Let $p$ be the partial pressure of  $CO_2$ and $H_2$ at equilibrium. The given reaction is-

$CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}+H_{2}(g)$
Initial concentration        4 bar               4 bar             0              0
At equilibrium                4 - $p$                 4 - $p$             $p$          $p$

Therefore, we can write,
$10.1= \frac{p_{CO_2}.p_{H_2}}{p_{CO}.p_{H_2O}}$
$\\10.1= \frac{p^2}{(4-p)^2}\\ 3.178 = \frac{p}{(4-p)}\\ p=(4-p)\times 3.178$
By solving the above equation we get,  $p$ = 3.04

Hence, the partial pressure of dihydrogen at equilibrium is 3.04 bar

If the value of $K_c$ is in the range of $10^{-3}$ to $10^{3}$, then the reaction has appreciable concentrations of both reactants and products.

Therefore, the third reaction (c)$(K_c=1.8)$ will have an appreciable concentration of reactants and products.

We have,

equilibrium constant of the reaction = $2\times 10^{-50}$
the concentration of dioxygen $[O_2]$ = $1.6 \times 10^{-2}$

the given reaction is-

$3O_{2}(g)\rightleftharpoons 2O_{3}(g)$
Then we have,
(equilibrium constant)
$K_c = [O_3 (g)]^2 / [O_2 (g)]^3$

$\\2\times 10^{-50} = [O_3]^2/(1.6\times 10^{-2})^3\\ {[O_3]^2} = 2\times 10^{-50}\times(1.6\times 10^{-2})^3$

$=8.192\times 10^{-56}$

${[O_3]}=\sqrt{8.192\times 10^{-56}} = 2.86 \times 10^{-28}$

Thus the concentration of dioxygen is $2.86 \times 10^{-28}$

Question 7.34(a)     The reaction,

$CO(g)+3H_{2}(g)\rightleftharpoons CH_{4}(g)+H_{2}O(g)$

is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of $CO$, 0.10 mol of $H_2$ and 0.02 mol of $H_2O$ and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Given that,
Total volume = 1L
0.3 mol of $CO$, 0.10 mol of dihydrogen($H_2$)and the 0.02 mol of water($H_2O$)
the equilibrium constant = 3.90

Let $x$ be the concentration of methane at equilibrium. The given reaction is-

$CO(g)+3H_{2}(g)\rightleftharpoons CH_{4}(g)+H_{2}O(g)$
At equilibrium,                    0.3 mol/L         0.1 mol/L        $x$           0.02 mol/L

Therefore,

$K_c = \frac{0.02\times x}{(0.3)(0.1)}=3.90$
$\Rightarrow x = \frac{(3.9)(0.3)(0.1)^3}{(0.02)}$
$=5.85\times 10^{-2}$

Thus the concentration of methane at equilibrium is $=5.85\times 10^{-2}$

$HNO_{2},CN^{-},HClO_{4},F^{-},OH^{-},CO_{3}^{-2} and S^{2-}$

A conjugate acid-base pair means that the species are differed by only one proton. for example; $HCl$ is an acid because it donates a proton to water. So $HCl-Cl^-$, and $H_3O^+-H_2O$these pairs are called conjugate acid-base pair.

 Species Conjugate acid-base $HNO_2$ $NO_2^{-}$(Base) $CN^-$ $HCN$(Acid) $HClO_4$ $ClO_4^-$(base) $F^-$ $HF$(acid) $OH^-$ $H_{2}O$(acid) $CO_3^{2-}$ $HCO_3^{-}$(acid) $S^{2-}$ $HS^{-}$(acid)

Lewis acid-
Those species which can accept the pair of electrons are called Lewis acids. For example  Boron trifluoride ($BF_3$), ammonium ion($NH_4^+$) and the hydrogen ion($H^+$). Among them water molecule is a Lewis base, it can donate pair of electrons.

When Brönsted acids lose their proton then they become a conjugate base of that corresponding acids.
Followings are the conjugate base of Brönsted acid-

• $HF-F^-$

• $H_2SO_4-HSO_{4}^-$

• $HCO_{3}^--CO_3^{2-}$

When Brönsted base accepts a proton then they become a conjugate acid of that corresponding base.
Followings are the conjugate acid of Brönsted base-

$NH_2^--NH_3$
$NH_3-NH_4^+$

$HCOO^--HCOOH$

When acid or base accept or lose a proton, they form conjugate acid or base of that corresponding species.

Lists of the conjugate acid and conjugate base of the given species-

 Species Conjugate acid Conjugate base $H_2O$ $H_3O^+$ $OH^-$ $HCO_{3}^-$ $H_2CO_{3}$ $CO_{3}^{2-}$ $HSO_{4}^-$ $H_2SO_{4}$ $SO_{4}^{2-}$ $NH_3$ $NH_4^+$ $NH_2^-$

Species which donate pair of an electron are called Lewis base and which accepts pair of electrons are called acid.

(a)  $OH^-$is a Lewis base since it can donate its lone pair of electrons.

(b) $F^{-}$ is a Lewis base since it can donate a pair of electrons.

(c)  $H^{+}$is a Lewis acid since it can accept a pair of electrons.

(d)  $BCl_{3}$ is a Lewis acid since it can accept a pair of electrons.

We have,
the concentration of Hydrogen ion sample is $3.8\times 10^{-3}$ M
So, $pH = -\log [H]^+$
$\\= -\log (3.8\times 10^{-3})\\ =3-\log 3.8\\ =2.42$

We have,
The pH of a sample of vinegar is 3.76
$pH$=?

Therefore,
$\\pH=-\log [H^+]\\ \log[H^+] = -pH$
Taking antilog on both sides we get,
$[H^+]$ = antilog (-3.76)
$[H^+]=1.74 \times 10^{-4}$

Hence the concentration of hydrogen ion $[H^+]=1.74 \times 10^{-4}$

We have,
IOnization constant of hydrogen fluoride, methanoic acid and hydrogen cyanide are $6.8\times 10^{-4},\ 1.8 \times 10^{-4}$ and $4.8\times 10^{-9}$ respectively.

It is known that,
$K_b = \frac{K_w}{K_a}$...........................(i)

$K_b$ of the conjugate base $F^-$
$=\frac{10^{-14}}{6.8 \times 10^{-4}}$

$=1.5 \times 10^{-11}$

Similarly,
By using the equation (i)
$K_b$ of the conjugate base $HCOO^-$
$=\frac{10^{-14}}{1.8\times 10^{-4}}$

$=5.6\times 10^{-11}$

Again, with the help of eq (i)
$K_b$ of the conjugate base $CN^-$
$=\frac{10^{-14}}{4.8\times 10^{-9}}$
$=2.8\times 10^{-6}$

We have,
The ionization constant of phenol is $1.0 \times 10^{-10}$, and
the concentration of phenol is 0.05 M
degree of ionisation = ?

Ionization of phenol;
$C_6H_5OH+H_2O\rightleftharpoons C_6H_5O^-+H_{3}O^+$
At equilibrium,
the concentration of various species are-
$[C_6H_5OH] = 0.05-x$
$[C_6H_5O^-] = [H_3O^+]=x$
As we see, the value of ionisation is very less. Also $x$ will be very small. Thus we can ignore $x$.

$\\K_a = \frac{[C_6H_5O^-][H_3O^+]}{[C_6H_5OH]}\\ 10^{-10}=\frac{x^2}{0.05}\\ x =\sqrt{10^{-10}\times 0.05}\\$

$x= 2.2\times 10^{-6}$

Hence the concentration of phenolate ion is $[C_6H_5O^-]= 2.2\times 10^{-6}$

Let $\alpha$ be the degree of dissociation of phenol in the presence of 0.01 M of phenolate ion.

$C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+$
Concentration                  (1 - $\alpha$) 0.05            0.05$\alpha$            0.05$\alpha$

So,
$[C_6H_5OH]= 0.05(1-\alpha) =0.05$
$[C_6H_5O^-]= 0.05\alpha+0.01 \approx 0.01M$

$[H_3O^+] = 0.05\alpha$

therefore,

$\\K_a =\frac{(0.01)(0.05\alpha)}{0.05}\\ 10^{-10}=0.01\alpha\\ \alpha = 10^{-8}$
The degree of dissociation is $10^{-8}$

We have,
1st ionisation constant of hydrogen sulphide is $9.1\times 10^{-8}$ and the 2nd dissociation constant is $1.2\times 10^{-13}$

Case 1st-(absence of hydrochloric acid)

To calculate the concentration of $HS^-$
Let $x$ be the concentration of $HS^-$and the ionisation of hydrogen sulphide is;
$H_2S\rightleftharpoons H^++HS^-$
0.1 M

At equilibrium, the concentration of various species are,
Since the dissociation constant is is very small. So, $x$ can be neglected.
the concentration of $H_2S$ = 1-$x$ M
the concentration of $HS^-$ and $H^+$ is $x$ M
So,
$K_a = \frac{x^2}{0.1}=9.1\times 10^{-8}$
from here $x$ can be calculated and we get,  $x = 9.54\times 10^{-5}M$

Case 2nd (In presence of 0.1 M, HCl)
Suppose $H_2S$ is dissociated is $x$ .Then at equilibrium,
$[H_2S] = 0.1-x\simeq 0.1,\ [H^+]=0.1+x\simeq 0.1$ and the $[HS^-] = x M$
So,
$K_a = \frac{x(0.1)}{0.1}=9.1\times 10^{-8}$
Thus the concentration of [$HS^-$] is $9.1\times 10^{-8}$

It is given,
The ionisation constant of acetic acid is $1.74\times 10^{-5}$ and concentration is 0.05 M

The ionisation of acetic acid is;

$CH_{3}COOH\rightleftharpoons CH_3COO^-+H^+$
Therefore,
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}= \frac{[H^+]^2}{[CH_3COOH]}$
$\therefore [H^+] = \sqrt{(1.74\times10^{-5})(5\times 10^{-2})} = 9.33\times 10^{-4}$$=[CH_3COO^-]$

So, the $pH$ of the solution = $-\log(9.33\times 10^{-4})$
=$4-\log(9.33)$
=$3.03$

We know that,
$\alpha = \sqrt{\frac{K_a}{C}}$
$\alpha = \sqrt{\frac{1.74\times10^{-5}}{0.05}} =1.86\times 10^{-2}$

We have,
pH of organic acid is 4.15 and its concentration is 0.01M

Suppose the organic acid be HA. The dissociation of organic acid can be written as;

$HA \rightleftharpoons H^+ + A^-$

$pH = 4.15$
$\\-\log[H^+] = 4.15\\ {[H^+]} = 7.08\times 10^{-5}$(By taking antilog of -4.15)

Now,
$K_a = [H^+] [A^-] / [HA]$

$[H^+] = [A^-] = 7.08\times10^{-5}$

[HA]  = 0.01

Then,

$\\K_a = (7.08\times10^{-5})^2 / 0.01\\ K_a = 5.01\times 10^{-7}$

Thus
$pK_a=-\log ( 5.01\times 10^{-7}) = 6.30$

(a)      0.003 M HCl

(b)      0.005 M NaOH

(c)      0.002 M HBr

(d)       0.002 M KOH

Assuming the complete dissociation. So, $\alpha =1$

(a) The ionisation of hydrochloric acid is
$HCl\rightleftharpoons H^++Cl^-$
Since it is fully ionised then $[H^+]=[Cl^-]=0.003M$
Therefore,
$pH$ of the solution$-\log (0.003)$
= $3-\log (3)$
= 2.52

(b) The ionisation of $0.005 M NaOH$
$NaOH\rightleftharpoons Na^++OH^-$
$[Na^+]=[OH^-]=0.005M$
Therefore,
$pOH$ of the solution =
$\\=-\log(0.005)\\ =3-\log5\\ =2.301$
$pH$ of the solution is equal to (14 - 2.301 =11.70)

(c)  The ionisation of $0.002 M HBr$
$HBr\rightleftharpoons H^++Br^-$
$[Br^-]=[H^+]=0.002M$
Therefore,
$pH$ of the solution =
$\\=-\log(0.002)\\ =3-\log2\\ =2.69$
$pH$ of the solution is equal to (2.69)

(d)  The ionisation of $0.002 M\ KOH$
$KOH\rightleftharpoons K^++OH^-$
$[OH^-]=[K^+]=0.002M$
Therefore,
$pOH$ of the solution =
$\\=-\log(0.002)\\ =3-\log2\\ =2.69$
$pH$ of the solution is equal to (14 - 2.69 = 11.31)

Question

2 g of TlOH dissolved in water to give 2 litre of solution.

Here, 2 g of $TlOH$ dissolves in water to give 2 litres of solution
So, the concentration of   $TlOH$ =
$[TlOH(aq)] = \frac{2}{2}g/L = \frac{1}{221}M$     (the molar mass of  $TlOH$ is 221)

$TlOH$ can be dissociated as $TlOH(aq)\rightarrow Tl^{+}+OH^-$

$OH^-(aq)= TlOH(aq)=\frac{1}{221}M$

Therefore, $K_w = [H^+][OH^-]$            (since Kw = $10^{-14}$)

So, the concentration of $[H^+]$ = $221\times 10^{-14}$

Thus $P^H = -\log[H^+] = -\log(221\times 10^{-14})$
= 11.65(approx)

Question

0.3 g of $Ca(OH)_2$ dissolved in water to give 500 mL of solution.

The calcium hydroxide ion dissociates into-
$Ca(OH)_{2}\rightarrow Ca^{2+}+2OH^-$
Molecular weight of $Ca(OH)_{2}$ = 74
the concentration of $[Ca(OH)_{2}]$ = $\frac{0.3\times 1000}{74\times500} = 0.0081 M$

$\therefore [OH^-]= [Ca(OH)_2] = 0.0081 M$

We know that,
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.0162}$
= $61.7 \times 10^{-14}$

Thus $P^H = - \log [H^+]= -\log (61.7 \times 10^{-14})$
= 14 - 1.79  = 12.21

Question

0.3 g of NaOH dissolved in water to give 200 mL of solution.

$NaOH$ dissociates into $NaOH\rightarrow Na^+ +OH^-$
So, the concentration of $[NaOH]$  = $\frac{0.3\times 1000}{200\times 40} M = 0.0375 M$

$\therefore [OH^-] = [NaOH] = 0.0375 M$

We know that ,
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.0375} = 2.66\times 10^{-13}$

Now, $P^H$ = $- \log[H^+]$
$\\=- \log(2.66\times 10^{-13})\\ =12.57$

Question

1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

We know that,
M1V1(before dilution) = M2V2(after dilution)

initially V1 = 1mL and M= 13.6 M
and V2 = 1L and M= ?

By putting all these values we get,

$M_2 = \frac{13.6\times 10^{-3}}{1}=1.36\times 10^{-2}$
$\therefore [H^+]=1.36\times 10^{-2}$

Thus $P^H = -\log [H^+] = -\log(1.36\times 10^{-2})$
= 1.86 (approx)

We have,

Degree of ionization(a) = 0.132
Concentration of bromoacetic acid (C) =  0.1 M

Thus the concentration of   $H_{3}O^+ = C.a$
= 0.0132

Therefore
$p^H = -\log[H^+] = -\log(0.0132)$
= 1.879

Now, we know that,
$K_a = C.a^2$
So, $pK_a = -\log(C.a^2)$
$\\=-\log (0.1\times (0.0132)^2)\\ =-\log (0.0017)\\ =2.76$(approx)

We have,
C = 0.005 M
$P^H$ = 9.95 and $P^{OH}$ = 14 - 9.95 = 4.05

we know that $P^{OH} = -\log[OH^-]$
$4.05 = -\log [OH^-]$

By taking antilog on both sides we get.

concentration of $[OH^-]$ = $8.91 \times 10^{-5}$

C.a = $8.91 \times 10^{-5}$
So, a = $1.782\times 10^{-2}$

We know that,
$K_b = C.a^2$
$\\= 0.005 \times (1.782\times 10^{-2})^2\\ =0.0158\times 10^{-4}$

Thus
$pK_b = -\log (K_b)$
$\\= -\log (0.0158\times 10^{-4})\\ =5.80$

We have,
C = 0.001 M
$K_b$ = $4.27\times 10^{-10}$
Degree of inozation of aniline (a) = ?
Ionization constant of the conjugate acid ($K_a$) = ?

We know that
$K_b$ = $C.a^2$
$4.27\times 10^{-10}$     =  (0.001)$a^2$
Thus  $a={\sqrt{4270\times 10^{-10}}}$
= $65.34\times 10^{-4}$

Then [Base] = C.a = ($65.34\times 10^{-4}$)($0.001$)
= $0.653\times 10^{-5}$

Now, $P^{OH} = -\log (0.65\times 10^{-5})=6.187$
$P^H = 14-P^{OH}$
= 14 - 6.187
= 7.813

It is known that,

$K_a\times K_b = K_w$
So, $K_a = \frac{10^{-14}}{4.27\times10^{-10}}$
$=2.34\times 10^{-5}$   This is ionization constant.

0.01M

We have,
C = 0.05 M
$pK_a = 4.74 = -\log(K_a)$
By taking antilog on both sides we get,
$(K_a) = 1.82\times 10^{-5} = C.(\alpha)^2$
from here we get the value of $\alpha= \sqrt{\frac{1.82\times 10^{-5}}{5\times10^{-2}}}$
$= 1.908\times 10^{-2}$

After adding hydrochloric acid, the concentration of $H^+$ ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease.

(i) when 0.01 HCl is taken
$CH_{3}COOH\rightleftharpoons CH_{3}COO^- +H^+$
Initial conc.                  0.05                     0                              0
after dissociation        0.05 - $x$               0.001+$x$                 $x$

As the dissociation is very small.
So we can write 0.001+$x$  $\approx$ 0.001 and  0.05 - $x$  $\approx$ 0.05

Now, $K_a= \frac{[CH_{3}COO^-][H^+]}{[CH_{3}COOH]}$
$\\=\frac{(0.001)(x)}{(0.05)}\\ =\frac{x}{50}$

So, the value of $x$ = $\frac{(1.82\times10^{-5})(0.05)}{0.01}$

Now degree of dissociation  = (amount dissociated) $/$(amount taken)

= $\frac{1.82\times 10^{-3}(0.05)}{0.05}$
= $1.82\times 10^{-3}$

0.1M in HCl ?

Let the $x$ amount of acetic acid is dissociated in this case

$CH_{3}COOH\rightleftharpoons CH_{3}COO^- +H^+$

Initial conc.                  0.05                     0                              0
after dissociation        0.05 - $x$               0.1+$x$                 $x$

As the dissociation is very small.
So we can write 0.1+$x$  $\approx$ 0.1 and  0.05 - $x$  $\approx$ 0.05

$K_a= \frac{[CH_{3}COO^-][H^+]}{[CH_{3}COOH]}$
$\\=\frac{(0.1)(x)}{(0.05)}\\ =2x$

So, the value of $x$ = $\frac{(1.82\times10^{-4})(0.05)}{0.1}$

Now the degree of dissociation  = (amount dissociated) $/$(amount is taken)

= $\frac{1.82\times 10^{-4}(0.05)}{0.05}$
= $1.82\times 10^{-4}$

We have,

(Degree of ionization) $K_b = 5.4\times 10^{-4}$
Concentration of dimethylamine = 0.02 M

$\therefore \alpha = \sqrt{\frac{K_b}{C}}= \sqrt{\frac{5.4\times 10^{-4}}{0.02}}$
$\alpha = 0.1643$

If we add 0.1 M of sodium hydroxide. It is a strong base so, it goes complete ionization

$NaOH\rightleftharpoons Na^++OH^-$
(0.1 M)      (0.1 M)

and also,

$(CH_{3})_2NH+H_{2}O\rightleftharpoons (CH_{3})_2NH^+_2+OH^-$
0.02- $x$                                           $x$                       $x$

$[OH^-]=x+0.1\approx 0.1$ (since the dissociation is very small)

Therefore,

$K_b = \frac{x(0.1)}{0.02}$
$x=5.4\times 10^{-4}(\frac{0.02}{0.1})$
$=0.0054$

Hence in the presece of 0.1 M of sodium hydroxide , 0.54% of dimethylamine get dissociated.

Human muscle-fluid, 6.83

We have $P^H =$6.83

It is known that $P^H = -\log[H^+]$

Therefore,
$6.83= -\log[H^+]$
By taking antilog on both sides we get,
$[H^+]= 1.48\times 10^{-7}M$

Human stomach fluid, 1.2

We have $P^H =$ 1.2

It is known that $P^H = -\log[H^+]$

Therefore,
$1.2= -\log[H^+]$
By taking antilog on both sides we get,
$[H^+]= 0.063\ M$

Human blood, 7.38

we have $P^H =$ 7.38

It is known that $P^H = -\log[H^+]$

Therefore,
$7.38 = -\log[H^+]$
By taking antilog on both sides we get,
$[H^+]= 4.17\times 10^{-8}\ M$

Human saliva, 6.4.

we have $P^H =$ 6.4

It is known that $P^H = -\log[H^+]$

Therefore,
$6.4 = -\log[H^+]$
By taking antilog on both sides we get,
$[H^+]= 3.98\times 10^{-7}\ M$

We already know that $p^H$ can be calculated as- $-\log[H^+]$
to calculate the concentration of $[H^+]$ = antilog (-$p^H$)

Thus, the hydrogen ion concentration of followings $p^H$ values are-

(i)  $p^H$of milk = 6.8
Since, $p^H=-\log[H^+]$
6.8 = $-\log[H^+]$
$\log[H^+]$ = -6.8

$[H^+]$ = anitlog(-6.8)

= $1.5\times10^{-7}M$

(ii) $p^H$of black coffee = 5.0

Since, $p^H=-\log[H^+]$

5.0 =$-\log[H^+]$

$\log[H^+]$= -5.0

$[H^+]$= anitlog(-5.0)

= $10^{-5}M$

(iii) $p^H$ of tomato juice = 4.2

Since, $p^H=-\log[H^+]$

4.2 =$-\log[H^+]$

$\log[H^+]$= -4.2

$[H^+]$= anitlog(-4.2)

= $6.31\times10^{-5}M$

(iv) $p^H$ of lemon juice = 2.2

Since, $p^H=-\log[H^+]$

2.2 = $-\log[H^+]$

$\log[H^+]$= -2.2

$[H^+]$= anitlog(-2.2)

=$6.31\times10^{-3}M$

(v) $p^H$ of egg white = 7.8

Since, $p^H=-\log[H^+]$

7.8 = $-\log[H^+]$

$\log[H^+]$ = -7.8

$[H^+]$= anitlog(-7.8)

= $1.58\times10^{-8}M$

We have 0.562 g of potassium hydroxide ($KOH$). On dissolving in water gives 200 mL of solution.
Therefore, concentration of $[KOH(aq)]$ = $\frac{0.561\times 1000}{200}g/L$

= 2.805 g/L

$=2.805 \times \frac{1}{56.11} M=0.05M$
$KOH(aq) \rightarrow K^+(aq) + OH^-(aq)$
It is a strong base. So, that it goes complete dissociation.

$[OH^-] = 0.05M = [K^+]$

It is known that,

$\\K_w = [H^+][OH^-]\\ {[H^+]}=\frac{K_w}{[OH^-]}$
$=\frac{10^{-14}}{0.05} = 2\times 10^{-13}M$

Therefore,
$p^H = -\log (2\times 10^{-13})=12.69$

By given abova data, we know the solubility of $Sr(OH)_2$ at 298 K = 19.23 g/L

So, concentration of $[Sr(OH)_2]$

= $19.23 / 121.63 M$ (Molecular weight of $Sr(OH)_2$ = 121.63 u)

= 0.1581 M

$Sr(OH)_2(aq) \rightarrow Sr^{2+}(aq) + 2 (OH^-)(aq)$

$\therefore Sr^{2+} = 0.1581M$
and the concentration of $[OH^-]$$= 2 \times 0.1581M = 0.3162 M$
Now

It is known that,
$K_w = [OH^-] [H^+]$

$[H^+] = \frac{10^{-14}}{0.3162}$
= $3.16 \times 10^{-14}$

Therefore $p^H = -\log(3.16 \times 10^{-14}) =13.5$

Let the degree of ionization of propanoic acid be $\alpha$. Then Let suppose we can write propanoic acid to be HA,

It is known that,

We have
ionization constant of propanoic acid ($K_a$)= $1.32\times 10^{-5}$ and the concentration is 0.005 M

$\alpha = \sqrt{\frac{K_a}{C}}$
By putting the values in above formula we get,

$\alpha = \sqrt{\frac{1.32\times 10^{-5}}{0.05}} = 1.62\times 10^{-2}$
[Acid] = $[H_{3}O]^+$ = C.$\alpha$ = $8.15\times 10^{-4}$

Therefore, $p^H = -\log [H_{3}O^+]$
$\\= -\log [8.15\times 10^{-4}]$
$=3.08$

If we add 0.01M hydrochloric acid then,

$AH+H_{2}O\rightleftharpoons H_{3}O^+ +A^-$
initial con                     C                               0                 0
at equi.                     C - $x$ $\approx$ C                    0.01 + $x$            $x$

Now, by using the formula of $K_a = \frac{(x)(0.01)}{C-x}$
$=\frac{(x)(0.01)}{C}$

The value of $x$ is calculated as  ;
$\Rightarrow 1.32\times 10^{-5}\times \frac{0.01}{0.01} = 1.32 \times 10^{-5}$  (Degree of ionisation)

We have,
Concentration of cyanic acid = 0.1 M
$p^H=-\log [H^+]$ $= 2.34$

Therefore, the concentration of $[H^+]$ = antilog (-2.34)
= $4.5 \times 10^{-3}$

It is known that,
$[H^+]$ $=C.\alpha$  = $4.5 \times 10^{-3}$
$\alpha = \frac{4.5\times 10^{-3}}{0.1} = 4.5\times 10^{-2}$

Then Ionization constant ($K_a$) $=C.\alpha^2$ =
$=(0.1)(4.5\times10^{-2})^2$
$=2.02\times 10^{-4}$

We have,
Ionization constant of nitrous acid = $4.5\times 10^{-5}$
Concentration of sodium nitrite ($NaNO_2$) = 0.04 M
Degree of hydrolysis can be calculated as;
$K_h = \frac{K_w}{K_a}=\frac{10^{-14}}{4.5\times 10^{-4}} = 0.22\times 10^{-10}$
Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid ($HNO_2$)
$NO_2^-+H_{2}O\rightleftharpoons HNO_{2}+OH^-$
Suppose $x$ moles of salt undergoes hydrolysis, then the concentration of-
$[NO_2^-]=0.04-x \approx 0.04$
$[HNO_2^-]=x$, and
$[OH^-] = x$

Therefore
$k_h = \frac{x^2}{0.04}=0.22\times 10^{-10}$
from here we can calculate the value of $x$ ;
$\Rightarrow x = \sqrt{0.0088\times 10^{-10}} = 0.093 \times 10^{-5} = [OH^-]$

$\Rightarrow [H_3O^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.093\times 10^{-5}}$
$=10.75 \times 10^{-9} M$

Now $p^H = -\log (10.75 \times 10^{-9} M) = 7.96$

Therefore the degree of hydrolysis

Given,
$p^H$  =  3.44

We know that

$p^H=-\log [H^+]$
By taking antilog on both sides we get,
$[H^+]$ = antilog (- 3.44)

$\therefore [H^+] = 3.63\times 10^{-4}$

pyridinium hydrochloride completely ionised.

Then $K_h$ =  (conc. of products)/ (conc, of reactants)
=$\frac{ ( 3.63\times10^{-2} )^2 }{0.02}$        (? Concentration is 0.02M)

$\Rightarrow K_h = 6.58 \times 10^-6$

Now,
$K_h = K_w / K_a$

$\Rightarrow Ka = K_w / K_h$

$= 10^{-14} / 6.58 \times 10^{-6} = 1.51 \times 10^{-9}$(approx)

Salts of strong acid and strong base are neutral in nature for example-

• $NaCl (NaOH + HCl)$

• $KBr(KOH+ HBr)$

Salts of a strong base and weak acid are basic in nature for example-

• $NaCN(HCN+NaOH)$

• $NaNO_{2}(HNO_{2}+NaOH)$

• $KF (KOH+HF)$

Salts of strong acid and a weak base are acidic in nature for example-

• $NH_4NO_{3} (NH_4OH+HNO_{2})$

We have,
Ionisation constant of chloroacetic acid($K_a$) is $1.35\times 10^{-3}$
The concentration of acid = 0.1 M
Ionisation if acid, =
$ClCH_2COOH\rightleftharpoons ClCH_2COO^-+H^+$

We know that,
$\Rightarrow K_a =\frac{[ClCH_2COO^-][H^+]}{[ClCH_2COOH]}$....................(i)
As it completely ionised
$[ClCH_2COO^-]=[H^+]$

Putting the values in eq (i)
$1.35\times 10^{-3} = \frac{[H^+]^2}{0.02}$
$[H^+] = \sqrt{1.35\times 10^{-3}\times 0.02}=1.16\times 10^{-2}$
Therefore, $pH$ of the solution = $-\log (1.16\times 10^{-2})$
= $2-\log(1.16)$
= $1.94$

Now,

0.1 M  $\dpi{100} ClCH_2COONa$ (sod. chloroacetate) is basic due to hydrolysis-

$\dpi{100} ClCH_2COO^-+ \: H_2O\rightleftharpoons CH_2ClCOOH+OH^-$

For a salt of strong base+strong acid

$\dpi{100} \\pH=7+\frac{pK_a+logC}{2}=7+\frac{2.87+log0.1}{2}\\pH=7.94$

We have the ionic product of water at 310 K is $2.7 \times 10^{-14}$
It is known that,
ionic product $K_w = [H^+][OH^-]$

SInce $[H^+]=[OH^-]$, therefore $K_w = [H^+]^2$

$\Rightarrow K_w$ at 310 K is $2.7 \times 10^{-14}$
$\therefore K_w = 2.7\times 10^{-14} = [H^+]^2$
here we can calculate the value of $[H^+]$ concentration.

$[H^+] = \sqrt{2.7 \times 10^{-14}} = 1.64 \times 10^{-7}$

Thus, $p^H = -\log[H^+]$
$= -\log(1.64\times 10^{-7})$
$= 6.78$

Hence the $p^H$ of neutral water is 6.78

Question

10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl

Given that,
Vol. of 0.2 M $Ca(OH)_2$ = 10 mL
Vol. of 0.1 M HCl = 25 mL

therefore, by using the formula,

$M(OH^-)= \frac{M_1V_1(base)-M_2V_2(acid)}{V_1+V_2}$
By substituting the value in these equations, we get;

$\\\Rightarrow \frac{(0.2 \times 2)- (0.1\times 2)}{10+25}\\ =\frac{1.5}{25}=0.06$

Now, $p^{OH} = -\log[OH^-]$
$\\= -\log(0.06)\\ =1.221$

since $p^H+p^{OH} = 14$
$p^H=14-p^{OH}$
= 14-1.221
=   12.78

Question

10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2

In this case, both the solutions have the same number of moles of $H^+$and $OH^-$, therefore they both can get completely neutralised. Hence the  $pH$ = 7.0

Question

c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

Given that,

Volume of 0.1 M $KOH$ = 10 mL, and

Volume of 0.1 M $H_2SO_{4}$ = 10 mL

So, by using the formula of,

$M(H^+)=\frac{M_1V_1(acid)-M_2V_2(base)}{V_1+V_2}$
By putting the values we get,

$\\\Rightarrow \frac{2(0.1\times 10)-0.1\times 10}{10+10}\\ =\frac{1}{20}\\ =5\times 10^{-2}$

Hence, $pH = -\log[H^+] = -\log(5\times 10^{-2})= 1.30$

Solubility product is the product of ionic concentrations in a saturated solution.
$K_{sp}=[A^+][B^-]$
(i) silver chromate ($Ag_2CrO_{4}$)
Ionization of silver chromate

$Ag_2CrO_{4}\rightleftharpoons 2Ag^++CrO_{4}^{2-}$
Let "$s$" be the solubility of $Ag_2CrO_{4}$
$[Ag^+] = 2s$
$[CrO_{4}^{2-}] = s$
According to the table $K_{sp}$ of $Ag_2CrO_{4}$ = $1.1\times 10^{-12}$

$\\\Rightarrow 1.1\times 10^{-12} = (2s)^2.s\\ =1.1\times 10^{-12} =2s^3$

$s = \sqrt[3]{\frac{1.1\times 10^{-12}}{4}}$
$=0.65 \times 10^{-4}$

(ii) Barium chromate ($BaCrO_{4}$)
Ionization of silver chromate

$BaCrO_{4}\rightleftharpoons Ba^{2+}+CrO_{4}^{2-}$
Let "$s$" be the solubility of $BaCrO_{4}$
$[Ba^{2+}] = s$
$[CrO_{4}^{2-}] = s$
According to the table