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:

Options

A. 5
B. 9
C. 7
D. 8

Answer

B

Explanation

`HA +NaOH hArrNaA +H_(2)O`
`K=10^(9)`
`:.` For `NaA +H_(2)O hArr NaOH +HA,K=10^(-9)`
Initial 0.1 M
At eqm `(0.1 -alpha)M " "alpha M " "alphaM`
`([NaOH][HA])/([NaA])=10^(-9)`
`(alpha^(2))/(0.1-alpha)=10^(-9)`
As `alpha` is very small , `0.1 -alpha ~~0.1`
`alpha^(2)=10^(-9)xx0.1`
`alpha=10^(-5)`
`[OH^(-)]=[NaOH]=10^(-5)M` ltbr. `pH =14 0pOH =14-5=9`