Solveeit Logo
Login

Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of
[CH3CH2COO−][CH3CH2COOH]\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH] }
required to make buffer is _____.
Given : Ka(CH3CH2COOH) = 1.3 × 10–5

A

0.03

B

0.13

C

0.23

D

0.33

Answer

0.13

Explanation

Solution

The correct answer is (B) : 0.13
CH3CH2COOH⇌CH3CH2COO−+H+CH_3CH_2COOH⇌CH_3CH_2COO^−+H^+
From Henderson equation
pH=pKa+log[CH3CH2COO−][CH3CH2COOH]pH=pK_a+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}
4=−log1.3×10−5log[CH3CH2COO−][CH3CH2COOH]4=−log 1.3×10^{−5} log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}
−log10−4=−log1.3×10−5+log[CH3CH2COO−][CH3CH2COOH]−log^{10^{−4}}=−log1.3×10^{−5}+log\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}
−log10−4=−log1.3×10−5[CH3CH2COOH][CH3CH2COO−]−log^{10^{−4}}=−log1.3×10^{−5} \frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}
10−4=1.3×10−5[CH3CH2COOH][CH3CH2COO−]10^{−4}=1.3×10^{−5}\frac{[CH_3CH_2COOH]}{[CH_3CH_2COO^-]}
[CH3CH2COO−][CH3CH2COOH]=0.13\frac{[CH_3CH_2COO^−]}{[CH_3CH_2COOH]}=0.13