Verified Solution • 2025–26 Session moderate

QJEEM22Ch1c63: JEE Main 2022 Chemistry - Class 11 JEEM22Ch1c63 Solution: A student needs to prepare a buffer solution of propanoic acid…

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH3CH2COO−][CH3CH2COOH]{{[C{H_3}C{H_2}CO{O^ - }]} \over {[C{H_3}C{H_2}COOH]}} required to make buffer is ___________.

Given : Ka(CH3CH2COOH)=1.3×10−5{K_a}(C{H_3}C{H_2}COOH) = 1.3 \times {10^{ - 5}}

Multiple Choice Options

  • A 0.03
  • B 0.13
  • C 0.23
  • D 0.33

Step-by-Step Verified Solution

CH3CH2COOH⇌CH3CH2COO−+H+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+}

From Henderson equation

pH=pKa+log⁡[CH3CH2COO−][CH3CH2COOH]\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}

4=−log⁡1.3×10−5+log⁡[CH3CH2COO−][CH3CH2COOH]4=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}

−log⁡10−4=−log⁡1.3×10−5+log⁡[CH3CH2COO−][CH3CH2COOH]-\log 10^{-4}=-\log 1.3 \times 10^{-5}+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}

−log⁡10−4=−log⁡1.3×10−5[CH3CH2COOH][CH3CH2COO−]-\log 10^{-4}=-\log 1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}

10−4=1.3×10−5[CH3CH2COOH][CH3CH2COO−]10^{-4}=1.3 \times 10^{-5} \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}

[CH3CH2COO−][CH3CH2COOH]=0.13\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=0.13

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