Calculate the degree of ionization of 0.01M solution of HCN.Ka = 4.8 X 10-10.Also calculate H^{+ }ion concentration of the solution

^{ }

^{ }

HCN + H

_{2}O $\iff $ H

_{3}O

^{+}+ CN

^{-}K

_{a}= 4.8 X 10

^{-10}

as water is present it also dissociates H

_{2}O + H

_{2}O $\iff $ H

_{3}O

^{+}+ OH

^{- }K

_{w}

_{ }= 1 X 10

^{-14}

The initial and equilibrium concentration of the reactants and products for HCN, H

_{3}O

^{+}and CN

^{-}will be

Initial conc 0.01M 0 0

Change -0.01α 0.01α 0.01α

Equilibrium conc 0.01-0.01α or 0.01(1-α) cα cα

As HCN is a weak acid as evidenced by low Ka value, α value is negligible and 1-α = 1 and 0.01(1-α) = 0.01

${K}_{a}=\frac{\left[H3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=\frac{(0.01\alpha )(0.01\alpha )}{0.01}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}4.8\times {10}^{-10}=0.01{\alpha}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\alpha}^{2}=(4.8\times {10}^{-10})(0.01)=4.8\times {10}^{-12}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\alpha =\sqrt{4.8\times {10}^{-12}}=2.2\times {10}^{-6}$

degree of dissociation is 2.2 X 10

^{-6}

For pH, we need the concentration of H

_{3}O

^{+ }which is contributed by HCN and H

_{2}O dissociation 9as HCn is a weak acid, this cannot be ignored)

[H

_{3}O

^{+}] i.e. [H

^{+}] form HCN dissociation = c$\alpha $ = 0.01 X ( 2.2 X 10

^{-6}) = 2.2 X 10

^{-8}M

[H

_{3}O

^{+}] i.e. [H

^{+}] form water dissociation = 1X 10

^{-7}M

Total [H

^{+}] = (2.2 X 10

^{-8}) + (1 X 10

^{-7}) = 1.22 X 10

^{-7}M

pH = -log [H+] = -log (1.22 X 10

^{-7}) = - (-6.9137) = 6.91.

**
**