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Question: What is the ratio of volumes of \[C{H_3}COOH(0.1N)\] to \[C{H_3}COONa(0.1N)\] required to prepare a ...

What is the ratio of volumes of CH3COOH(0.1N)C{H_3}COOH(0.1N) to CH3COONa(0.1N)C{H_3}COONa(0.1N) required to prepare a buffer solution of pH=5.74pH = 5.74? (Given pKaofCH3COOH=4.74p{K_a} of C{H_3}COOH = 4.74)
10:110:1
5:15:1
1:51:5
1:101:10

Explanation

Solution

The solution which maintains its pH fairly constant and is capable of resisting the change in pH in addition to small amounts of acid or base is called a buffer solution. Buffer action is the ability of a buffer solution to resist any change in pH on the addition of a small quantity of acid or base.

Complete step by step answer:
Step 1
A freshly prepared solution always has a definite pH value. It is usually unable to maintain the same pH value for a long time. The value changes either keeping it for a long time or by adding a small amount of acid or bases. Sometimes in the industrial or the biological process, it is required to keep a solution for a long time or on the addition of a small amount of acid and base the pH value should not change. That is the requirement of a buffer solution.
Step 2
There are two types of buffer solution:
Acidic buffer – the solution of a weak acid and its salt of a strong base is called an acidic buffer.
Basic buffer – the solution of a weak base and its salt of a strong acid is called a basic buffer.
Step 3
Henderson’s equation helps to calculate the pH of a buffer solution.
To find out the acidic buffer in the given question
and
CH3COONa(aq)→CH3COO−(aq)+Na+(aq)C{H_3}COONa(aq)\xrightarrow{{}}C{H_3}CO{O^ - }(aq) + N{a^ + }(aq)
The law of mass action and law of equilibrium suggests,
Ka=[H3O+][CH3COO−][CH3COOH]{K_a} = \dfrac{{[{H_3}{O^ + }][C{H_3}CO{O^ - }]}}{{[C{H_3}COOH]}}
[H3O+]=Ka[CH3COOH][CH3COO−][{H_3}{O^ + }] = {K_a}\dfrac{{[C{H_3}COOH]}}{{[C{H_3}CO{O^ - }]}}
Step 4
In the buffer solution due to the common ion effect the acid is negligible, thus concentration of the acid is taken as the initial concentration and the concentration of the acetate ion is taken as the initial concentration of the salt.
[H3O+]=Ka[Acid][Salt][{H_3}{O^ + }] = {K_a}\dfrac{{[Acid]}}{{[Salt]}}
Taking negative logarithm on both the sides
−log⁡[H3O+]=−log⁡Ka−log⁡[Acid][Salt]- \log [{H_3}{O^ + }] = - \log {K_a} - \log \dfrac{{[Acid]}}{{[Salt]}}
pH=pKa+log⁡[Salt][Acid]pH = p{K_a} + \log \dfrac{{[Salt]}}{{[Acid]}}
Let the volume of sodium acetate taken be V and the volume of acetic acid taken is V’
Therefore,
5.74=4.74+log⁡[0.1/V][0.1/V′]5.74 = 4.74 + \log \dfrac{{[0.1/V]}}{{[0.1/V']}}
log⁡[0.1/V][0.1/V′]=1\log \dfrac{{[0.1/V]}}{{[0.1/V']}} = 1
log⁡[V′][V]=log⁡10\log \dfrac{{[V']}}{{[V]}} = \log 10
Taking antilog on both the sides
V′V=10\dfrac{{V'}}{V} = 10

So, the correct answer is “Option A”.

Note: Human blood and seawater are examples of the natural buffer which maintains their pH by the complex buffer action of the various salts present in it.