`0.1M NaOH` is titrated with `0.1M HA` till the end point. `K_(a)` of HA is `5.6xx10^(-6)` and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ?

`0.1M NaOH` is titrated with `0.1M HA` till the end point. `K_(a)` of

1.	`0.1M NaOH` is titrated with `0.1M HA` till the end point. `K_(a)` of HA is `5.6xx10^(-6)` and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point ?
Answer» `{:(NaOH+,HArarr,NaA+,H_(2)O),(0.1,0.1,0,0),(0,0,0.1,0.1):}` `[NaA] = (0.1)/(2)` because volume is doubled on mixing. `NaA` is salt of weak acid `HA` and stron base `NaOH`. For such salt solution: `pH = 7 +(1)/(2) pK_(a) +(1)/(2)logC` `= 7+(1)/(2)(-log 5.6 xx 10^(-5))+(1)/(2)log((0.1)/(2))` `= 7+(1)/(2) (5-log 5.6) +(-0.65)` `= 7 +(5.25)/(2) - 0.65 = 8.975`