Equilibrium

NCERT Solution

Part 5

Question 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.

Answer: NaNO₂ is a salt of a strong base (NaOH) and a weak acid (HNO₂)

NO₂^- + H₂O ⇌ HNO₂ + OH^-

K_b `=([HN\O_2][OH^-])/([NO_2^-])`

Or, `(K_w)/(K_a) = (10^(-14))/(4.5xx10^(-4))``=0.22xx10^(-10)`

If x mole of salt undergo hydrolysis, then concentration of various species in solution will be as follows:

[NO₂^-] = 0.04 – x ∼ 0.04

[HNO₂] = [OH^-] = x

`K_b=(x^2)/(0.04)=0.22xx10^(-10)`

Or, `x^2=0.0088xx10^(-10)`

Or, `x=0.093xx10^(-5)`

So, [OH^-] = 0.093 × 10^-5 M

[H₃O⁺] = `(10^(-14))/(0.093xx10^(-5))=10.75xx10^(-9)`

pH = - log(10.75 × 10^-9) = 7.96

So, degree of hydrolysis `=x/(0.04)=(0.093xx10^(-5))/(0.04)`

= 2.325 × 10^-5

Question 62: A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Answer: pH = 3.44 = - log [H⁺]

Or, [H⁺] = 3.63 × 10^-4

Now, `K_b=((3.63xx10^(-4))^2)/(0.02)=6.6xx10^(-6)`

We know, `K_b=(K_w)/(K_a)`

So, `K_a=(K_w)/(K_b)`

`=(10^(-14))/(6.6xx10^(-6))=1.51xx10^(-7)`

Question 63: Predict is the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF.

Answer: NaCl + H₂O ⇌ NaOH + HCl

Here, NaOH is a strong base and HCl is a strong acid. So, a salt formed out of neutralization between them will be neutral salt.

KBr + H₂O ⇌ KOH + HBr

KBr is a neutral salt because it is made from strong base and strong acid.

NaCN + H₂O ⇌ NaOH + HCN

This salt is made from strong base and weak acid. So it is basic salt.

NH₄NO₃ + H₂O ⇌ NH₄OH + HNO₃

This salt is made from weak base and strong acid. So, it is acidic salt.

NaNO₂ + H₂O ⇌ NaOH + HNO₂

This salt is made from weak acid and strong base. So, it is basic salt.

KF + H₂O ⇌ NaOH + HF

This salt is made from weak acid and strong base. So, it is basic salt.

Question 64: The ionization constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Answer: `K_b=Cα^2`

So, `α=sqrt((K_a)/C)`

`=sqrt((1.35xx10^(-3))/(0.01))=0.116`

So, [H⁺] = Cα = 0.1 × 0.116 = 0.0116

pH = - log[H⁺] = 1.94

Sodium chloroacetate is a salt of weak acid (chloroacetic acid) and strong base (sodium hydroxide)

`K_b=([ClH_2COOH][OH^-])/([ClCH_2COO^-])`

Or, `K_b=(K_w)/(K_a)`

`=(10^(-14))/(1.35xx10^(-3)`

Also, `K_=(x^2)/(0.1)`

Or, 0.740 × 10^-11 `=(x^2)/(0.1)`

Or, `x^2=0.074xx10^(-11)=0.74xx10^(-12)`

Or, `x=0.86xx10^(-6)`

Or, [OH^-] = 0.86 × 10^-6

So, [H⁺ `=(K_w)/(0.86xx10^(-6))`

`=(10^(-4))/(0.86xx10^(-6))`

Or, [H⁺ = 1.162 × 10^-8

pH = - log [H⁺] = 7.94

Question 65: Ionic product of water at 310 K is 2.7 × 10^-14. What is the pH of neutral water at this temperature?

Answer: Ionic product of water K_w = [H⁺][OH^-]

Let us assume [H⁺] = x

As [H⁺] = [OH^-]

So, `K_w=x^2`

Or, `x^2=2.7xx10^(-14)`

Or, `x=1.64xx10^(-7)`

Now, pH = - log [H⁺]

= -log [1.64 × 10^-7] = 6.78

Question 66: Calculate the pH of the resultant mixtures:

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

Answer: No. of moles of H⁺ `=(25xx0.1)/(1000)=0.0025` mol

No. of moles of OH^- `=(10xx0.2xx2)/(1000)=0.0040` mol

So, the excess of OH^- = 0.0040 – 0.0025 = 0.0015

So, [OH^- `=(0.0015)/(35xx10^(-3))=0.428`

pOH = -log[OH]= 1.36

pH = 14 – 1.36 = 12.36 (Not matched)

(Note: It is a neutral mixture because Ca(OH)₂ is strong base while HCl is strong acid.

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

Answer: No. of moles of H⁺ `=(2xx10xx0.1)/(1000)=0.002` mol

No. of moles of OH^- `=(2xx10xx0.1)/(1000)=0.002` mol

As no. of moles for both H⁺ and OH^- is same, the solution is neural, i.e. pH = 7.

Answer: No. of moles of H₊ `=(2xx10xx0.1)/(1000)=0.002` mol

No. of moles of OH^- `=(10xx0.1)/(1000)=0.001` mol

Excess of H⁺ = 0.002 – 0.001 = 0.001 mole

So, [H⁺ `=(0.001)/(20xx10^(-3)=0.5`

pH = - log (0.05) = 1.30

Question 67: Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in following table. Determine also the molarities of individual ions.

Salt	K_sp
Silver chromate	1.1 × 10^-12
Barium chromate	1.2 × 10^-10
Ferric hydroxide	1.0 × 10^-38
Lead chloride	1.6 × 10^-5
Mercurous iodide	4.5 × 10^-29

Answer: Silver Chromate

Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄^-

So, K_sp = [Ag⁺]²[CrO₄^2-]

If solubility is s then

[Ag⁺] = 2s and [CrO₄^2-] = s

Or, `K_(sp)=(2s)^2.s=4s^3`

Or, `1.1xx10^(-12)=4s^3`

Or, `s^2=0.275xx10^(-12)`

Or, `s = 0.65xx10^(-4)` M

Molarity of Ag⁺ = 2s = 2 × 0.65 × 10^-4

= 1.30 × 10^-4 M

Molarity of CrO₄^- = s = 0.65 × 10^-4 M

Barium Chromate

K_sp = [Ba₂₊][CrO₄^2-]

If solubility of BaCrO_{4 is s}

Then [Ba²⁺] = [CrO₄^2-] = s

So, `K_sp=s^2`

Or, `s^2=1.2xx10^(-10)`

Or, `s=1.09xx10^(-5)`

Molarity of Ba₂₊ = Molarity of CrO₄²⁺

= s = 1.09 × 10^-5 M

Ferric Hydroxide

K_sp = [Fe²⁺][OH^-]³

If s is solubility of Fe(OH)₃

Then [Fe³⁺] = s and [OH^-] = 3s

So, `K_(sp)=s.(3s)^3=27s^4`

Or, `27s^4=1.0xx10^(-38)`

Or, `s^4=0.037xx10^(-38)`

Or, `s=1.39xx10^(-10)`

Molarity of Fe³⁺ = s = 1.39 × 10^-10 M

Molarity of OH^- = 3s = 4.17 × 10¹⁰ M

Lead Chloride

K_sp = [Pb²⁺][Cl^-]²

If s is the solubility of PbCl₂

Then [Pb²⁺] = s and [Cl^-] = 2s

So, `K_(sp)=s.(2s)^2`

Or, `4s^3=1.6xx10^(-5)`

Or, `s^3=0.4xx10^(-10)=4xx10^(-4)`

Or, `s=1.58xx10^(-2)` M

Molarity of Pb²⁺ = 1.58 × 10^-2 M

Molarity of Cl^- = 2s = 3.16 × 10^-2 M

Mercurous Iodide

K_sp = [Hg_g²⁺][I^-]²

If s is the solubility of mercurous iodide then

[Hg₂²⁺ = s and [I^-] = 2s

So, `K_(sp)=s.(2s)^2=4s^3=4.5xx10^(-29)`

Or, `s^3=1.125xx10^(-29)`

Or, `s=2.24xx10^(-10)` M

Molarity of H₂²⁺ = s = 2.24 × 10^-10 M

Molarity of I^- = 2s = 4.48 × 10^-10 M

Question 68: The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10^-12 and 5.0 × 10^-13 respectively. Calculate the ratio of molarities of their saturated solutions.

Answer: Ag₂CrO₄ ⇌ Ag²⁺ + 2CrO₄^2-

If s is the solubility of Ag₂CrO₄

Then,`K_(sp)=(2s)^2.s=4s^3`

Or, `4s^3=1.1xx10^(-12)`

Or, `s=6.5xx10^(-5)` M

AgBr ⇌ Ag⁺ + Br^-

If s' is the solubility of AgBr

Then, `K_(sp)=s'^2=5.0xx10^(-13)`

Or, `s=7.07xx10^(-7)` M

So, ratio of molarities of their saturated solution is

`(6.5xx10^(-5))/(7.07xx10^(-7))=91.9`

Question 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)

Answer: When we add equal volumes of sodium iodate and cupric chlroate solutions then molar concentrations of both the solutions are reduced to half, i.e. 0.001 M.

NaIO₃ → Na⁺ + IO₃^-

Cu(ClO₃)₂ → Cu²⁺ + 2ClO₃^-

Solubility equilibrium of copper iodate can be written as follows:

Cu(IO₃)₂ → Cu⁺ + 2IO₃^-

Ionic product = [Cu²⁺][IO₃^-]²

= 0.001 × 0.001² = 1 × 10^-4

Since ionic product is less than K_sp hence precipitation does not take place.

Question 70: The ionization constant of benzoic acid is 6.46 × 10^-5 and K_sp 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?

Answer: pH = 3.19 = - log [H⁺]

Or, [H⁺ = 6.46 × 10^-4 M

C₆H₅COOH ⇌ C₆H₅COO^- + H⁺`

K_a`=([C_6H_5CO\O^-][H^+])/([C_6H_5CO\OH]`

Or, `([C_6H_5CO\OH])/([C_6H_5CO\O^-])=([H^+])/(K_a)`

`=(6.46xx10^(-4))/(6.46xx10^(-5))=10`

Let us assume that solubility of C₆H₅COOAg is x mol L^-1

`[Ag^+]=x`

[C₆H₅COOH] + [C₆H₅COO^-] = x

Or, 10[C₆H₅COO^-] + [C₆H₅COO^-] = x

Or, [C₆H₅COO^-] = `x/(11)`

Or, 2.5 × 10^-13 = `(x^2)/(11)`

Or, `x=1.66xx10^(-6)` M L^-1

Solubility of silver benzoate in pH 3.19 solution is 1.66 × 10^-6 mol L^-1.

Now, let us assume that solubility of C₆H_{COOAg is x' mol L^-1}

Then , [Ag⁺ = x'M

[C₆H₅COO^-] = x'M

K^sp = (x')^2

Or, `x' =sqrt(K_(sp))`

`=sqrt(2.5xx10^(-13))=5xx10^(-7)` mol L^-1

So, `x/(x')=(1.66xx10^(-6))/(5xx10^(-7))=3.32`

So, C₆H₅COOAg is approximately 3.32 times more soluble in low pH solution.

Question 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)

Answer: Let us assume that maximum concentration of each solution is x mol L^-1. After mixing, the volume of the concentrations of each solution will be reduced to half, i.e. x/2.

So, [FeSO₄] = [Na₂S] `=x/2` M

Or, [Fe²⁺] = [S²⁼`=x/2` M

K_sp = `(x/2)^2`

Or, `(x^2)/4=6.3xx10^(-18)`

Or, `x=5.02xx10^(-9)`

If concentrations of both the solutions are equal to or less than 5.02 × 10^-9 there will be no precipitation of iron sulphide.

Question 72: What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate K^sp is 9.1 × 10^-6)

Answer: CaSO₄ ⇌ Ca²⁺ + SO₄^2-

K_sp = [Ca²⁺][SO₄^2-]

If solubility of CaSO₄ is s

Then, K_sp = s² = 9.1 × 10^-6

Or, s = 3.02 × 10^-3 mol L^-1

Molar mass of CaSO₄ = 136 g mol^-1

So, solubility of CaSO₄

= 3.02 × 10^-3 × 136 = 0.41 g L^-1

This means that we need 1 L water to dissolve 0.41 g of CaSO₄

So, to dissolved 1 g of CaSO₄, we need 1/(0.41)=2.44 L of water

Question 73: The concentration of sulphide ion in 0.1 M 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: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂, in which of these solutions precipitation will take place?

Answer: Before mixing:

[S^2- = 1.0 × 10^-19 M (volume = 10 mL)

[M²⁺] = 0.04 M (volume = 5 mL)

After mixing: [S^2-] = ? and [M²⁺

Volume becomes 15 mL for each, after mixing

[S^2-] `=(1.0xx10^(-19)xx10)/(15)=6.67xx10^(-20)` M

[M²⁺] `= (0.04xx5)/(15)=1.33xx10^(-2)` M

Ionic Product = [M²⁺][S^2-]

`= (1.33xx10^(-2))(6.67xx10^(-20))`

= 8.87 × 10^-22

Ionic product is greater than K^sp of Zn and CdS. So, precipitation will happen.