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#### The equilibrium constant at 298 K for a reaction A+BC+D is 100.  If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L−1) will be : Option: 1  0.182   Option: 2  0.818   Option: 3  0.818   Option: 4  1.182

As we dicussed in the concept

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

are equilibrium concentration

For reaction

Initial concentration IM      IM       IM       IM

At equilibarium if degree of dissociation is then

1-   1-      H   1+

9=

Concentration of D is 1+=I+

=1.818

#### 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.  If pKb of ammonia solution is 4.75, the pH of the mixture will be : Option: 1 3.75 Option: 2 4.75 Option: 3 8.25 Option: 4 9.25

Initially, before reaction

mmol of $\mathrm{NH_3}$ = 10

mmol of $\mathrm{HCl}$ = 5

Therefore, after reaction,

mmol of $\mathrm{NH_3}$ = 5

mmol of $\mathrm{NH_4Cl}$ = 5

Now, putting the values of concentrations in the equation

$\mathrm{pOH = pK_b + \log \frac{[Salt]}{[Base]}}$

$\Rightarrow \mathrm{pOH = pK_b \ \ ( \because [Salt]=[Base])}$$\Rightarrow \mathrm{pOH = 4.75}$

Thus, we have

$\Rightarrow \mathrm{pH =14-pOH =14- 4.75=9.25}$

Hence, the correct answer is Option (4)

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#### Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6.  If ionisation constant of HA is 10−5, the ratio of salt to acid concentration in the buffer solution will be : Option: 1  4 : 5 Option: 2 1 : 10 Option: 3  10 : 1 Option: 4  5 : 4

As we learnt,

$\mathrm{pH=pKa+\log \frac{[\text {salt}]}{[\text {acid}]}}$

$\therefore \mathrm{6=5+\log \frac{[\text {salt}]}{[\text {acid}]}}$

$\Rightarrow \mathrm{\log \frac{[\text {salt}]}{[\text {acid}]}=1}$

$\Rightarrow \mathrm{\frac{[\text {salt}]}{[\text {acid}]}=10}$

Hence, the correct answer is Option (3)

#### Consider the following statements: (a) The pH of a mixture containing 400mL of 0.1 M H2SO4 and 400ml of 0.1M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with $K_{a}=10^{-5}$ has a pH=5. The degree of dissociation of this acid is 50%. (d) The Le Chatelier's principle is not applicable to common-ion effect. The correct statements are : Option: 1 (a), (b) and (d) Option: 2 (a), (b) and (c) Option: 3 (b) and (c) Option: 4 (a) and (b)

(a) $H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

Mole of H2SO4 = 400 x 0.1 = 40

Mole of NaOH = 400 x 0.1 = 40

for 40 mol of H2SO4 , Need 80 mol of NaOH. But NaOH is 40 mol only.

So, NaOH is a limiting reagent.

Hence, 40 moles of NaOH will react with only 20 moles of H2SO4.

$\mathrm{[H^{+}]=\frac{20\times 2}{800(\textup{Total volume})}=\frac{1}{20}}$

$\mathrm{pH=-\log \left (\frac{1}{20} \right )=1.3}$

(b) $\mathrm{\log \frac{K_{w2}}{K_{w1}}=\frac{\Delta H}{2.303R}\left [\frac{1}{T_1}-\frac{1}{T_2} \right ]}$

Thus, the ionic product of water is temperature-dependent.

(c) $\mathrm{K_a=10^{-5}\textup{ and }pH=5\Rightarrow [H^{+}]=10^{-5}}$

Now,

$\\\mathrm{K_a=\frac{C\alpha ^{2}}{(1-\alpha )}}\\\\\\ \mathrm{K_a=\frac{[H^{+}]\alpha }{(1-\alpha )}}$

$10^{-5}=\frac{10^{-5}\alpha }{(1-\alpha )}$

$\mathrm{\alpha =50\%\textup{ or 1/2}}$

(d) Le-chatelier's principle is applicable to the common ion effect.

So, statement (d) is wrong.

Therefore, option(2) is correct.

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#### Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is_________: Option: 1 pH = 10.6 Option: 2 pH = 12.7 Option: 3 pH = 9.8 Option: 4 pH = 11

For given solutions, we have:

Moles of NaOH  4/40 = 0.1 moles

Moles of H2SO4 = 9.8/98 = 0.1 moles

Molarity of NaOH = 0.1/100L
And molarity of H2SO4 = 0.1/100

Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:

Total volume = 50L

Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04

Milliequivalents of H2SO4 = 10x(0.2/100)x2 = 0.02

Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02

[OH-] = 4x10-4

pOH = -log[4x10-4]

pOH = -log4 - log10-4

pOH = -0.60 + 4 = 3.4

Further, we know:

pH = 14 - 3.4

pH = 10.6

Hence, the option number (1) is correct.

#### For the following Assertion and Reason, the correct option is : Assertion : The pH of water increases with increase in temperature. Reason :    The dissociation of water into  and  is an exothermic                     reaction. Option: 1 Assertion is not true, but reason is true. Option: 2 Both assertion and reason are true, and the reason is the correct explanation for the assertion. Option: 3 Both assertion and reason are false. Option: 4 Both assertion and reason are true, but the reason is not the correct explanation for the assertion.

The pH of water does not increase with the increase in temperature.
The dissociation of water molecules into ions is bond breaking and is, therefore, an endothermic process.

Therefore, Option(3) is correct.

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#### 3g of acetic acid is added to 250 mL of 0.1M HCl and the solution made up to 500 mL. To 20 mL of this solution $\frac{1}{2}$ mL. of 5 M NaOH is added. The pH of the solution is_______. [ Given: pKa of acetic acid = 4.75, molar mass of acertic acid = 60 g/mol, log3 = 0.4771] Neglect any chnages in volume.

Given,

m mole of acetic acid in 20 mL = 2
m mole of HCl in 20 mL = 1
m mole of NaOH = 2.5

$\mathrm{CH_{3}COOH\: +\: NaOH(remaining)\: \rightarrow \: CH_{3}COONa\: +\:water}$
2                              3/2                                      0                                 0
0.5                             0                                       3/2

$\mathrm{pH\: =\: pK_{a}\: +\: log\frac{\frac{3}{2}}{2}}$

pH = 4.74 + log 3

pH = 4.74 + 0.48 = 5.22

Thus, the correct answer is 5.22

Hence, the option number (1) is correct.

#### Arrange the following solution in the decreasing order of $pOH$ A) $0.01M \; HCl$ B)  C)  D) $0.01M \; NaCl$Option: 1Option: 2Option: 3Option: 4

$pOH + pH = 14$

If $pH$ is low(means more acidic) then $pOH$ is high.

Thus, acidic order: $HCl > NaCl > CH_{3}COONa > NaOH.$

Then the order of $pOH$ will be:

$A > D > C > B.$

Therefore, Option(3) is correct.

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#### For the reaction: Option: 1Option: 2Option: 3Option: 4

The reaction occurs as follows:

$\mathrm{Fe_{2}N(s)\: +\: \frac{3}{2}H_{2}(g)\: \rightleftharpoons \: 2Fe(s)\: +\: NH_{3}(g)}$

$\mathrm{\Delta n\: =\: -\frac{1}{2}}$

We know that:

$\\\mathrm{k_{p}\: =\: k_{c}(RT)^{\Delta n}}\\\\\mathrm{Thus,\: k_{p}\: =\: k_{c}(RT)^{-\frac{1}{2}}}\\\\\mathrm{k_{c}\: =\: k_{p}(RT)^{\frac{1}{2}}}$

Therefore, Option(2) is correct.

#### If the solubility product of $AB_{2}$ is  $3.20\times 10^{-11}M^{3},$ then the solubility of $AB_{2}$ in pure water is ________$\times10^{-4} mol L^{-1}.$ [Assuming that neither kind of ion reacts with water]

If Solubility is s then

Ans = 2

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