Question

A solution of weak acid HA was titrated with base NaOH. The equivalent point was reached when 40 mL. Of 0.1 M NaOH has been added. Now 20 mL of 0.1 M HCl were added to titrated solution, the pH was found to be 5.0 What will be the pH of the solution obtained by mixing 20 mL of 0.2 M NaOH and 20 mL of 0.2 M HA?
A. `7`
B. `9`
C. `10`
D. `11`

Answer 1

Correct Answer - B
`HA + NaOH rarr NaA + H_(2)O` milli moles of salt
`NaA` or `A^(-) = 40 xx 0.1 = 4`
`{:("Now",A^(-)+,H^(+)rarrHA),("Initial milli moles",4,2),("Final milli moles",2,-2):}`
Acidic buffer solution is formed and `[A^(-)] = [HA]`
`pH = pK_(a) + "log" ([A^(-)])/([HA]) rArr pK_(a) = 5`
Now `HA + NaOH rarr NaA + H_(2)O`, hydrolysis of
`A^(-)` will takes place
`[NaA] = ("milli moles of acid")/("total volume") `
`= (20 xx 0.2)/(20 + 20) = 0.1`
`pH = 1/2 (pK_(w) + pK_(a) + "log" C) = (1)/(2) [14+5-1] = 9`