Ionic EquilibriumHard
Question
50 mL of 2N acetic acid mixed with 10 mL of 1N sodium acetate solution will have an approximate pH of
(Ka = 10-5)
(Ka = 10-5)
Options
A.4
B.5
C.6
D.7
Solution
Meq. of acetic acid = 50 × 2 = 100
Meq. of CH3COONa = 10 × 1 = 10
pH = - log Ka + log
or pH = - log 10-5 + log 
= 4
Meq. of CH3COONa = 10 × 1 = 10
pH = - log Ka + log
or pH = - log 10-5 + log = 4Create a free account to view solution
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