Ionic EquilibriumHard
Question
To a 200 ml of 0.1 M weak acid HA solution 90 ml of 0.1 M solution of NaOH be added. Now, what volume of 0.1 M NaOH be added into above solution so that pH of resulting solution be 5. [(Ka(HA) = 10-5]
Options
A.2 ml
B.20 ml
C.10 ml
D.15 ml
Solution
HA + NAOH NaA + H2O
t = 0 20 9 0 0
t = t 11 0 9 0
t = 50% 10 0 10
10 ml NaOH is required.
t = 0 20 9 0 0
t = t 11 0 9 0
t = 50% 10 0 10
10 ml NaOH is required.
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