JEE Main 2025 · Chemistry · Ionic Equilibrium · medium

HCN has Ka=1×10^-9. 55 mL of 0.2 M NaCN is mixed with 10 mL of 0.1 M HCl. The pH of the resulting solution is (nearest integer)

Numerical answer

\mathrm{HCN}

has

K_a=1\times10^{-9}

55\ \mathrm{mL}

0.2\ \mathrm{M}

\mathrm{NaCN}

is mixed with

10\ \mathrm{mL}

0.1\ \mathrm{M}

\mathrm{HCl}

. The pH of the resulting solution is (nearest integer)

Show answer and solution

Answer: 10

Moles:

\mathrm{CN^-}=0.055\times0.2=0.011

;

\mathrm{HCl}=0.010\times0.1=0.001

\mathrm{HCl}

converts

0.001

mol of

\mathrm{CN^-}

\mathrm{HCN}

, leaving

\mathrm{CN^-}=0.010

mol and

\mathrm{HCN}=0.001

mol: a buffer.

\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{[\mathrm{CN^-}]}{[\mathrm{HCN}]}=9+\log10=10

(the volume cancels in the ratio).

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