Question

$K_a$ for $HCN$ and ${CH}_{3}COOH$ are $4.9\times {10}^{-10}$ and $1.8\times {10}^{-5}$ respectively. Calculate the equilibrium constant for the reaction:

${CH}_{3}COOH+NaCN\rightleftharpoons {CH}_{3}COONa+HCN$

• A. $3.674\times {10}^3$
• B. $3.674\times {10}^4$
• C. $4.674\times {10}^4$
• D. None of these

Answer 1

The correct option is B $3.674\times {10}^4$

$C{H}_{3}COOH+NaCN\rightleftharpoons C{H}_{3}COONa+HCN$

$Ka=\frac{\left[HCN\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]\left[C{N}^{-}\right]}$ ----- (a)

$K{a}_{\left(HCN\right)}=\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}$ -------- (1)

$K{a}_{\left(C{H}_{3}COOH\right)}=\frac{\left[H^{+}\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{H}_{3}COOH\right]}$ ----- (2)

$Ka=\frac{K{a}_{\left(C{H}_{3}COOH\right)}}{K{a}_{\left(HCN\right)}}$

$Ka=\frac{1.8\times {10}^{-5}}{4.9\times {10}^{-10}}$

$Ka=3.674\times {10}^4$