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

`HA+NaOH rarr NaA+H_(2)O`
or `HA+OH^(-)rarr A^(-)+H_(2)O`
`K_(eq.)=10^(9)=([A^(-)][H_(2)O])/([HA][OH^(-)])`
Also `HAhArrH^(+)+A^(-)`
`K_(a)= ([H^(+)][A^(-)])/([HA])`
`:. (K_(eq.))/(K_(a))=(1)/(K_(w))`
or `K_(a)=10^(9)xx10^(-14)=10^(-5)`
Thus, for `A^(-)H_(2)OhArrHA+OH`
`[OH^(-)]= c.h=c. sqrt((K_(H))/(c ))= sqrt((K_(w).c)/(K_(a)))`
`=sqrt((10^(-14)xx-0.1)/(10^(-5)))=10^(-5)`
`:. [H^(+)]=10^(-9) implies pH =9`