The pH of basic buffer mixtures is given by : pH = pK a + log 
, whereas pH of acidic buffer mixtures is given by : pH = pK a + log 
. Addition of little acid or base although shows no appreciable change in pH for all practical purposes, but since the ratio 
for 
changes, a slight decrease or increase in pH results.
(i) A solution containing 0.2 mole of dichloroacetic acid (K a = 5 × 10 –2 ) and 0.1 mole sodium dichloroacetate in one litre solution has [H + ]
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) Since CHCl 2 COOH is relatively strongs acid havings more K a .
CHCl 2 .COONa 
CHCl 2 COO – + Na +
0.1 0.1
CHCl 2 .COOH 
CHCl 2 COO – + H +
0.2 0 0
(0.2–x) (x+0.1) x
∴ K a = 
or 5 × 10 –2 = 
.
∴ x = 0.05.
(ii) Let V mL of NaOH be needed to give CH 3 COONa.
NaOH + CH 3 COOH 
CH 3 COONa + H 2 O
0.2×V 50×0.2 0 0
— [10–0.2V] 0.2V 0.2V
∴ pH = p K a + log 
= pK w – pK b + log 
= 14 – 9.26 + log 
= 14 – 9.26 + log 
4.74 = 4.74 + log 
∴ V = 
= 25 mL.
(iii) CH 3 COONa + HCl → CH 3 COOH + NaCl
1 1 0 0
0 0 1 1
∴ [CH 3 COOH] = 
= 1.
∴ [H + ] = C α = C 
= 
= 
or pH 1 = – 
logK a = 
pK a .
CH 3 COOH + CH 3 COONa
1 1
∴ pH = pK a + log 
.
pH 2 = pK a .
∴ 
= 
.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems