Integer Answer Type Questions for JEE: Equilibrium | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. A solution of HCOOH is titrated with KOH solution. When half of the acid is neutralised, AgCN is added to make a saturated solution. Calculate the solubility of AgCN in milligram per 100 litres of solution. K_a (HCN) = 10^–10, K_sp (AgCN) = 10^–16, Ka (HCOOH) = 10^–4.

Ans. 134 mg / 100 L
pH of buffer = pK_a + log [Salt/Acid] = 4 + log 1/1 = 4
AgCN ⇌ Ag⁺ +  CN^– x is solubility of AgCN

Hence,

⇒ [CN^–] = 10^–6x = 10^–16
K_sp = x × 10^–6x
⇒ x² = 10^–10 ⇒ x = 10^–5 M
Hence, the solubility of AgCN is 10^–5 M = 134 mg / 100 L.

Q.2. What volume in ml of 0.10 M sodium formate solution should be added to 50 ml of 0.05 M formic acid to produce a buffer solution of pH 4.1 pK_a for formic acid is 3.80.

Ans. 50 ml.
Let x ml of 0.10 M sodium formate be added.
∴ Moles of HCOONa added = 0.1x × 10^-3
Moles of HCOOH = 0.05 × 50 × 10^-3 = 2.5 × 10^-3

For acidic buffer, pH = pK_a + log [salt/acid]
4.1 = 3.8 + log 0.04x.
On solving, x = 50 ml.

Q.3. 500 ml of 0.150 M AgNO₃ solution is mixed with 500 ml of 1.09 M Fe²⁺ solution and the reaction is allowed to reach equilibrium at 25°C.
Ag⁺(aq) + Fe²⁺(aq) ⇌ Fe³⁺(aq)  +  Ag(s)
For 25 ml of the equilibrium solution, 30 ml of 0.0833 M KMnO₄ were required for oxidation. Calculate the approximate equilibrium constant for the reaction at 25°C.

Ans. 3.0
Ag+(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)
No. of m moles initially 500 × 0.15 500 × 1.09 0 0
= 75 =545
M moles at equilib. 75 -x 545 - x x x
On titration of reaction mixture with KMnO₄, only Fe²⁺ reacts with it.
∴ Equivalents of Fe²⁺ in 25 ml = Equivalents of KMnO₄

= 30 × 10^-3 × 0.0833 × 5

545 -x = 500;
x = 45

Q.4. An unknown volume and unknown concentration of weak acid HX is titrated with NaOH of unknown concentration. After addition of 10.0 cm³ of NaOH solution, pH of solution is 5.7 and after the addition of 20.0 cm³ of NaOH solution, the pH is 6.3. Calculate the pK_a for the weak acid, HX. (Given: antilog of 0.6 ≈ 4)

Ans. 6.0
Let the molarity and volume of HX be M₁ and V₁ ml respectively while the molarity of NaOH be M₂.

of 10 c.c of NaOH
Since, weak acid HX and NaX are left after the reaction they will constitute an acidic buffer.
 …(1)
Let V₂ ml be the volume of NaOH required to neutralize given HX completely.
∴ At equivalence point, M₁V₁ = M₂V₂

Dividing the numerator and denominator of log term by M₂, we get

Similarly, after the addition of 20 c.c of NaOH, we have

Subtracting equation (2) from equation (3)

Taking antilog,

∴ V₂ = 30 ml.
Putting V₂ in equation (2),

pK_a = 5.7 + log 2 = 5.7 + 0.3 = 6.0.

Q.5. Calculate the weight in mg of HCl added to 100 ml of 0.1 N BOH to have its pH = 6.6 and K_b = 6.25 × 10^-8 ; Antilog (-7.4) = 3.98 × 10^-8. (Assume there is no change in volume on addition of HCl)

Ans. 223 mg.
Let the mass of HCl added be x mg
BOH + HCl → BCl + H₂O

This constitutes basic buffer. For basic buffer, the [OH^-] is given by

On solving, n = 223
∴ Weight of HCl required = 223 mg.