If the equilibrium constant of the reaction of weak acid HA with strong base is 109, then pH of 0.1 M Na A is

HA+OH−⇌H2O+A− K=A−[HA]OH− HA⇌H++A− Ka=H+A−[HA] ∴ KaK=H+OH−=Kw Ka=KwK=10−14×109=10−5 ∴ pKa=5 A- solution is alkaline due to hydrolysis, ∴ pH=7+pKa2+logC2 =7+52+log0.12=9