Question

pH of a mixture of $ 1M $ benzoic acid $ \left( {p{K_a} = 4.20} \right) $ and $ 1M $ $ {C_6}{H_6}COONa $ is $ 4.5 $ , what is the volume of benzoic acid required to prepare a $ 300mL $ buffer $ \left[ {\log 2 = 0.3} \right] $ ?

Answer 1

Hint: Buffer solution is that solution, whose pH value remains relatively constant upon addition of a small amount of acid or base. They resist the change in its pH value upon adding acid or base. There are three types of buffer solution: Acid buffer, Base buffer, Salt buffer.
Acid buffers are formed by a mixture of a weak acid and its salt with a strong base.
$ pH = pKa + \log \dfrac{{{{\left[ {anion} \right]}_{salt}}}}{{{{\left[ {acid} \right]}_{\alpha \to 0}}}} $

Complete answer:
Benzoic acid is a weak acid and the salt sodium benzoate is benzoic acid’s salt with a strong base. The mixture of these forms an acid buffer.
It is given in the question:
pH of a mixture of $ 1M $ benzoic acid $ \left( {p{K_a} = 4.20} \right) $ and $ 1M $ $ {C_6}{H_6}COONa $ is $ 4.5 $
To find: the volume of benzoic acid required to prepare a $ 300mL $ buffer solution.
Substituting the known values in the formula above and simplifying:
 $ 4.5 = 4.20 + \log \dfrac{{{V_{salt}}}}{{{V_{acid}}}} $
 $
  0.3 = \log \dfrac{{{V_{salt}}}}{{{V_{acid}}}} \\
  \log 2 = \log \dfrac{{{V_{salt}}}}{{{V_{acid}}}} \\
  $
Substitute $ \log 2 = 0.3 $
 $
  2 = \dfrac{{{V_{salt}}}}{{{V_{acid}}}} \\
  2{V_{acid}} = {V_{salt}} \\
  $
Also the volume of the buffer is $ 300mL $ then the sum of the volume of salt and acid will be equal to the volume of the buffer.
 $ {V_{salt}} + {V_{acid}} = 300 $
Solving the equations by substituting the value of $ 2{V_{acid}} = {V_{salt}} $ in the above equation:
 $
  2{V_{acid}} + {V_{acid}} = 300 \\
  3{V_{acid}} = 300 \\
  {V_{acid}} = 100mL \\
  $
And by equation $ 2{V_{acid}} = {V_{salt}} $
 $
  {V_{salt}} = 2\left( {100} \right) \\
  {V_{salt}} = 200mL \\
  $
Thus, the volume of benzoic acid required to prepare a $ 300mL $ buffer solution of benzoic acid and sodium benzoate is $ 100mL $.

Note:
When a small amount of acid is added to the acid buffer the hydrogen ion is used by the salt to again form the acid. And when base is added the acid of the buffer solution uses it to form anion and water.
$ C{H_3}CO{O^ - } + {H^ + } \to C{H_3}COOH \\
  C{H_3}COOH + O{H^ - } \to C{H_3}CO{O^ - } + {H_2}O $
 In this way the addition of a small amount of acid and base does not affect the pH of buffer solutions.