Ionic EquilibriumHard
Question
A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) has a pH of 5. At the end point, the volume of same base required is 27 mL. Ka of acid is (log2 = 0.3)
Options
A.1.8 × 10-5
B.8 × 10-6
C.1.8 × 10-6
D.8 × 10-5
Solution
m. moles of HA taken = 27 × 0.1 = 2.7
HA + OH- A- + H2O
t=0 2.7 1.2
teq 1.5 - 1.2
pH = pKa + log
⇒ 5 = pKa + log 
= pKa + log
∴ pKa = 5.1 ⇒ Ka = 8 × 10-6.
HA + OH- A- + H2O
t=0 2.7 1.2
teq 1.5 - 1.2
pH = pKa + log
⇒ 5 = pKa + log = pKa + log∴ pKa = 5.1 ⇒ Ka = 8 × 10-6.
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