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Chemistry Question on Equilibrium

The dissociation constants for acetic acid and HCNHCN at 25∘C25^{\circ} C are 1.5×10−51.5 \times 10^{-5} and 4.5×10−104.5 \times 10^{-10} respectively. The equilibrium constant for the equilibrium CN−+CH3COOH⇌HCN+CH3COO−CN ^{-}+ CH _{3} COOH \rightleftharpoons HCN + CH _{3} COO ^{-} would be :

A

3.0×1053.0 \times 10^{5}

B

3.0×10−53.0 \times 10^{-5}

C

3.0×10−43.0 \times 10^{-4}

D

3.0×1043.0 \times 10^{4}

Answer

3.0×1043.0 \times 10^{4}

Explanation

Solution

Kc=Ka(CH3COOH)×1Ka(HCN)K _{ c } = K _{ a \left( CH _{3} COOH \right)} \times \frac{1}{ K _{ a ( HCN )}}
=1.5×10−5×14.5×10−10=1.5 \times 10^{-5} \times \frac{1}{4.5 \times 10^{-10}}
≅3×104\cong 3 \times 10^{4}