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  • A student needs to prepare a buffer solution of propanoic acid and its sodium salt with . The ratio of <

A student needs to prepare a buffer solution of propanoic acid and its sodium salt with \mathrm{pH}= 4.

The ratio of \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]} required to make buffer is_________.

\text { Given : } \mathrm{K}_{\mathrm{a}}\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right)=1.3 \times 10^{-5}

Option: 1

0.03


Option: 2

0.13


Option: 3

0.23


Option: 4

0.33


Answers (1)

best_answer

Given,

\mathrm{pH=4, \mathrm{Ka}\left(\mathrm{CH}_{3} \mathrm{C} \mathrm{H}_{2} \mathrm{COOH}\right)=1.3 \times 10^{-5}}

We need to find out the ratio of \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}

 

As we know

\mathrm{pH=p{K a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}}

4=5-\log 1 . 3+\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}

\log \frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=\log 1.3-1=\log \frac{1.3}{10}

comparing Both sides, we have 

\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]}=\frac{1.3}{10}=0.13

Hence, Option(2) is correct.

Posted by

Deependra Verma

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