Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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).

Sagot :

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

https://brainly.com/question/14782315

#SPJ4