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Chemistry Question on Equilibrium

Ionisation constant of CH3COOHCH_3COOH is 1.7×10−51.7 \times 10^{-5} and concentration of H+H^+ ions is 3.4×10−43.4 \times 10^{-4} . Then, find out initial concentration of CH3COOHCH_3COOH molecules.

A

3.4×10−43.4 \times 10^{-4}

B

3.4×10−33.4 \times 10^{-3}

C

6.8×10−46.8 \times 10^{-4}

D

6.8×10−36.8 \times 10^{-3}

Answer

6.8×10−36.8 \times 10^{-3}

Explanation

Solution

CH3COOH⇌CH3COO−+H+CH_3COOH \rightleftharpoons CH_3COO^- + H^+
\hspace20mm K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}
Given that,
\hspace10mm [CH_3COO^-] = [H^+] = 3.4 \times 10^{-4} M
\hspace20mm K_a for CH3COOH=1.7×10−5CH_3COOH = 1.7 \times 10^{-5}
CH3COOHCH_3COOH is weak acid, so in it [CH3COOH][CH_3COOH] is equal to initial concentration. Hence,
\hspace5mm 1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4} )(3.4 \times 10^{-4} )}{[CH_4COOH]}
[CH3COOH]=3.4×10−4×3.4×10−41.7×10−5=6.8×10−3M[CH_3COOH] = \frac{3.4 \times 10^{-4} \times 3.4 \times 10^{-4}}{1.7 \times 10^{-5}} = 6.8 \times 10^{-3} M