**7.73 ** The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{-19} M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: in which of these solutions precipitation will take place?

We have,
the concentration of and the volume of the solution containing sulphur ion = 10 mL.
Volume of metal salts solution added = 5mL
Before mixing,
M
After mixing,
Volume = 15 mL
So, the concentration of
concentration of
So, the ionic product =
...

**7.72 ** What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_{sp} is 9.1 × 10^{-6})

We have,
The solubility product of calcium sulphate is .
given mass of calcium sulphate = 1g
Ionization of calcium sulphate;
Therefore,
Let the solubility of calcium sulphate be .
Then,
mol/L
Thus,
mass/ (mol. wt) volume Molarity
mass =
So, that to dissolve 1 g of calcium sulphate we need = of water.

**7.71** What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_{sp} = 6.3 × 10^{-18}).

We have,
The solubility product of the
Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume.
Let be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half.
So,
The ionisation of ferrous sulphide;
Therefore, for no precipitation, ionic product = solubility product
By...

**7.70 ** The ionization constant of benzoic acid is 6.46 × 10^{-5} and Ksp for silver benzoate is 2.5 × 10^{-13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

It is given that,
of buffer solution is 3.19. So, the concentration of ion can be calculated as;
= antilog (-3.19)
=
Ionization of benzoic acid;
It is given that
Therefore,
Let the

**7.69** Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_{sp} = 7.4 × 10^{-8} ).

We have,
solubility product () of cupric iodate =
When equal volumes of sodium iodate and cupric chlorate are mixed together the molar concentration of both the solution becomes half (= 0.001)
Ionization of cupric iodate is;
0.001 M 0.001 M
So, can be calculated as;
Sinc eionic product is less than the so no precipitation occurs.

**7.68 ** The solubility product constant of Ag_{2}CrO_{4} and AgBr is 1.1 × 10^{-12} and 5.0 × 10^{-13} respectively. Calculate the ratio of the molarities of their saturated solutions.

silver chromate ()
Ionization of silver chromate
Let "" be the solubility of
of =
Ionization of Silver bromide ()
of =
Now, the ratio of solubilities

**7.67** Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table. Determine also the molarities of individual ions.

Solubility product is the product of ionic concentrations in a saturated solution.
(i) silver chromate ()
Ionization of silver chromate
Let "" be the solubility of
According to the table of =
(ii) Barium chromate ()
Ionization of silver chromate
Let "" be the solubility of
According to the table of =
(iii) Ferric hydroxide ()
Ionization of Ferric...

**7.66** Calculate the pH of the resultant mixtures:

c) 10 mL of 0.1M H_{2}SO_{4} + 10 mL of 0.1M KOH

Given that,
Volume of 0.1 M = 10 mL, and
Volume of 0.1 M = 10 mL
So, by using the formula of,
By putting the values we get,
Hence,

**7.66** Calculate the pH of the resultant mixtures:

b) 10 mL of 0.01M H_{2}SO_{4} + 10 mL of 0.01M Ca(OH)_{2}

In this case, both the solutions have the same number of moles of and , therefore they both can get completely neutralised. Hence the = 7.0

**7.66 ** Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)_{2} + 25 mL of 0.1M HCl

Given that,
Vol. of 0.2 M = 10 mL
Vol. of 0.1 M HCl = 25 mL
therefore, by using the formula,
By substituting the value in these equations, we get;
Now,
since
= 14-1.221
= 12.78

**7.65 ** Ionic product of water at 310 K is 2.7 × 10^{-14}. What is the pH of neutral water at this temperature?

We have the ionic product of water at 310 K is
It is known that,
ionic product
SInce , therefore
at 310 K is
here we can calculate the value of concentration.
Thus,
Hence the of neutral water is 6.78

**7.64 ** The ionization constant of chloroacetic acid is 1.35 × 10^{-3}. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

We have,
Ionisation constant of chloroacetic acid() is
The concentration of acid = 0.1 M
Ionisation if acid, =
We know that,
....................(i)
As it completely ionised
Putting the values in eq (i)
Therefore, of the solution =
=
=
Now,
0.1 M (sod. chloroacetate) is basic...

**7.63 ** Predict if the solutions of the following salts are neutral, acidic or basic:

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

**7.62** A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Given,
= 3.44
We know that
By taking antilog on both sides we get,
= antilog (- 3.44)
pyridinium hydrochloride completely ionised.
Then = (conc. of products)/ (conc, of reactants)
= (? Concentration is 0.02M)
Now,
(approx)

**7.61** The ionization constant of nitrous acid is 4.5 × 10^{-4}. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

We have,
Ionization constant of nitrous acid =
Concentration of sodium nitrite () = 0.04 M
Degree of hydrolysis can be calculated as;
Sodium nitrite is a salt of sodium hydroxide (strong base) and the weak acid ()
Suppose moles of salt undergoes hydrolysis, then the concentration of-
, and
Therefore
from here we can calculate the value of ;
...

**7.60** The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

We have,
Concentration of cyanic acid = 0.1 M
Therefore, the concentration of = antilog (-2.34)
=
It is known that,
=
Then Ionization constant () =

**7.59** The ionization constant of propanoic acid is . Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Let the degree of ionization of propanoic acid be . Then Let suppose we can write propanoic acid to be HA,
It is known that,
We have
ionization constant of propanoic acid ()= and the concentration is 0.005 M
By putting the values in above formula we get,
[Acid] = = C. =
Therefore,
If we add 0.01M hydrochloric acid then,
...

**7.58 ** The solubility of at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

By given abova data, we know the solubility of at 298 K = 19.23 g/L
So, concentration of
= (Molecular weight of = 121.63 u)
= 0.1581 M
and the concentration of
Now
It is known that,
=
Therefore

**7.57 ** If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions.What is its pH?

We have 0.562 g of potassium hydroxide (). On dissolving in water gives 200 mL of solution.
Therefore, concentration of =
= 2.805 g/L
It is a strong base. So, that it goes complete...

**7.56** The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each

We already know that can be calculated as-
to calculate the concentration of = antilog (-)
Thus, the hydrogen ion concentration of followings values are-
(i) of milk = 6.8
Since,
6.8 =
= -6.8
= anitlog(-6.8)
=
(ii) of black coffee = 5.0
Since,
5.0 =
= -5.0
= anitlog(-5.0)
=
(iii) of tomato juice = 4.2
Since,
4.2 =
= -4.2
= anitlog(-4.2)
=
(iv) of lemon juice =...

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