Ionic EquilibriumHard
Question
If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 M HCOOH. the pH of the resulting solution is:
(Ka = 1.8 × 10-4, log 18 = 1.26)
(Ka = 1.8 × 10-4, log 18 = 1.26)
Options
A.3.74
B.5.64
C.7.57
D.3.42
Solution
HCOOH + KOH → HCOOK + H2O
milimole 20 10
10 - 10
pH = pKa + log
= 3.74 + log 
⇒ pH = 3.74
milimole 20 10
10 - 10
pH = pKa + log
= 3.74 + log ⇒ pH = 3.74Create a free account to view solution
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