Question

The equilibrium constant `(K_(c))` for the reaction of a weak acid `HA` with strong base `NaOH`is `10` at `25^(@)C`. Which of the following are correct deduction
A. The ionization constant `K_(a)` at `25^(@)C` is `10^(-5)`
B. pH of a `0.01 M` aqueous solution of `HA `at `25^(@)C` will be `3.5`
C. pH of a `0.01 M` aqueous solution of `NaA` at `25^(@)C` will be `9`.
D. If `K_(b)` of weak base `BOH` is `10^(-4)` at `25^(@)C`, equilibrium constant for neutralization of HA with BOH at `25^(@)C` will be `10^(5)`

Answer 1

Correct Answer - A::B::C::D
`underset(("Weak"))(HA) + NaOH hArr NaA + H_(2)O , K = 10^(9)`
`K = 1/(K_(h)) hArr = 10^(-9)`
A) `K_(h) = (K_(w))/(K_(a))` or `K_(a) = (K_(w))/(K_(h)) = 10^(-5)`
B) `pH` of `0.01 M HA = 1/2 pK_(a) - 1/2 "log" C = 3.5`
C) pH of `0.1 M aq.NaA = 7 + 1/2 pK_(a) + 1/2 "log" C = 9`
`K = (1)/(K_(h)) = (K_(a) xx K_(b))/(K_(w)) = (10^(-5) xx 10^(-4))/(10^(-14)) = 10^(5)`