Ionic EquilibriumHard
Question
A weak acid, HA is found to be 10% ionized in 0.01 M aqueous solution. Calculate thepH of a solution which is 0.1 M in HA and 0.05 M in NaA.
Options
A.5.365
B.6.355
C.3.653
D.6.593
Solution
α = 0.1,
Ka =
= 1.11 × 10-4
Now pKa = - log 1.11 × 10-4 = 3.9542
pH = pKa + log
= 3.9542 + log
= 3.65
Ka =
= 1.11 × 10-4Now pKa = - log 1.11 × 10-4 = 3.9542
pH = pKa + log
= 3.9542 + log
= 3.65Create a free account to view solution
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