Question

Question 7.66 Calculate the pH of the resultant mixtures:

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

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

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

Class XI Equilibrium Page 230

Answer 1

(a) Here, for base
M1 = [OH-] = 2 × 0.2 = 0.4 M
V1 = 10 ml so, M1V1 = 0.4 × 10 = 4 Mml
for acid,
M2 = [H+] = 0.1 M
V2 = 25 ml so, M2V2 = 0.1 × 25ml = 2.5 Mml

here, strength of base (M1V1) > strength of acid (M2V2) so, solution is basic .

so, [OH-] = {M1V1 - M2V2}/(V1 + V2)
= (4 - 2.5)/(10 + 25) = 0.043 M
now, use Arrhenius formula,
POH = -log[OH-] = -log(0.043 ) = 2 - 0.6335 = 1.3665
so, PH = 14 - POH = 14 - 1.3665 = 12.6335

similarly we can solve next all questions
(b) for acid ( H2SO4),
M1 = [H+] = 2 × 0.01M = 0.02 M , V1 = 10 ml
so, M1V1 = 0.02 × 10 = 0.2 Mml

for base Ca(OH)2,
M2 =[ OH-] = 2 × 0.01 M = 0.02M , V1 = 10 ml
so, M2V2 = 0.02M × 10ml = 0.2 Mml

because strength of H2SO4 = strength of base = 0.2 Mml
so, solution is neutral .
hence, pH of solution is 7 .

(c) for acid (H2SO4),
M1 = [H+] = 2 × 0.1 M = 0.2 M ,V1 = 10ml
M1V1 = 0.2 M × 10 ml = 2Mml

for base (KOH),
M2 = 0.1M , V2 = 10ml
so, M2V2 = 0.1M × 10 Mml= 1 Mml
because strength of acid (H2SO4) > strength of base ( KOH)
so, solution is acidic.
now, [H+] = (M1V1 - M2V2)/(V1+V2)
= (2 - 1)/(10+10) = 0.05 M

pH = -log[H+] = -log(5 × 10^-2 ) = 2 - 0.6990
= 1.301 ≈ 1.3