If the equilibrium constant for the reaction of weak acid HA with strong base is `10^(9)`, then pH of `0.1M` Na A is:

Answer

9

Explanation

`K_("neu.") = (1)/(K_(h)) rArr K_(h) = 10^(-9)` and
`HCO_(3)^(-) + H_(2)O hArr CO_(3)^(2-) + H_(3)O^(+), pK_(a2) = 10.25`
`K_(h) = (K_(w))/(K_(a)) rArr K_(a) = (10^(-14))/(10^(-9)) = 10^(-5)`
`rArr pKa = 5`
`pH = 7 + 1/2 pKa + 1/2 "log" C`
`rArr 7 + 5/2 - (0.1)/(2) = 9`