Respuesta :
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 +[tex]H_{2} O[/tex] → [tex]H^{+} + CN^{-}[/tex]
C(1-x) Cx Cx
Where x , is the degree of hydrolysis and
[tex]K_{h}[/tex] = C[tex]x^{2}[/tex]/(1-x)
We know that [tex]K_{h}[/tex] = [tex]K_{w}/K_{a}[/tex] = 1 ×[tex]10^{-14}[/tex]/4. 9 ×[tex]10^{-10}[/tex] = 2.04×[tex]10^{-5}[/tex]
[tex]K_{h}[/tex] = C[tex]x^{2}[/tex] = 2.04×[tex]10^{-5}[/tex] = 0.11 M×[tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 2.04×[tex]10^{-5}[/tex]/0.11 M
x = 1.36×[tex]10^{-2}[/tex]
[tex][OH^{-} ][/tex] = Cx = 1.36×[tex]10^{-2}[/tex] × 0.11 M = 0.15×[tex]10^{-2}[/tex]
[tex][H^{+} ][/tex] = 1 ×[tex]10^{-14}[/tex]/ 0.15×[tex]10^{-2}[/tex] = 6.66×[tex]10^{-12}[/tex]
pH = -㏒[tex][H^{+} ][/tex] = -㏒6.66×[tex]10^{-12}[/tex] = 11.17
The pH at the equivalence point for the titration is 11.17.
To learn more about equivalence point
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