Question

Calculate the ph at the equivalence point for the titration of 0. 22 m hcn with 0. 22 m naoh. (ka = 4. 9 × 10^–10 for HCN).

Answer 1

The pH at the equivalence point for the titration of 0. 22 m HCN with 0. 22 m NaOH is 11.17

Calculation,

Concentration of NaCN = 0. 22 m/ 2 = 0.11 M ( at equal volumes of acid and base will be used).

The equilibrium is ,

HCN +$H_{2} O$  → $H^{+} + CN^{-}$

C(1-x)              Cx      Cx

Where x , is the degree of hydrolysis and

$K_{h}$ = C$x^{2}$/(1-x)

We know that $K_{h}$ = $K_{w}/K_{a}$ = 1 ×$10^{-14}$/4. 9 ×$10^{-10}$ =  2.04×$10^{-5}$

$K_{h}$ =  C$x^{2}$ =   2.04×$10^{-5}$  = 0.11 M×$x^{2}$

$x^{2}$ =  2.04×$10^{-5}$/0.11 M

x = 1.36×$10^{-2}$

$[OH^{-} ]$ = Cx =  1.36×$10^{-2}$ × 0.11 M = 0.15×$10^{-2}$

$[H^{+} ]$ =  1 ×$10^{-14}$/ 0.15×$10^{-2}$ = 6.66×$10^{-12}$

pH = -㏒$[H^{+} ]$ = -㏒6.66×$10^{-12}$ = 11.17

The pH at the equivalence point for the titration is  11.17.

To learn more about  equivalence point

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