Question

$ 100mL $ of $ 0.02M $ benzoic acid $ \left( p{{K}_{a}}=4.2 \right) $ is titrated using $ 0.02M $ $ NaOH. $ pH after $ 50mi $ , and $ 100mL $ of $ NaOH $ have been added are
(A) $ 3.50,7 $
(B) $ 4.2,7 $
(C) $ 4.2,8.1 $
(D) $ 4.2,8.25 $

Answer 1

Hint :A mixture of acetic acid and sodium acetate act as a buffer solution and since in this reaction sodium hydroxide is reacting with acetic acid producing sodium acetate. Therefore, the overall solution is acting as a buffer in each other’s presence.

Complete Step By Step Answer:
The overall reaction when benzoic acid is reacting with sodium hydroxide is shown as:
 $ C6H5COO{{H}_{s}}+NaOH\to C6H5COONa_{_{aq}}^{+}+{{H}_{2}}{{O}_{aq}} $
Therefore, moles of are $ 0.02 $ moles and moles of is $ 0.02 $ moles. It shows that they are added in $ 1:1 $ mole ratios in the equation. Therefore, no moles are formed of moles. No of moles of used will be $ 0.02 $ moles as $ 0.02 $ mole of reacts with $ 0.02 $ mole of to give $ 0.02 $ mole. Since, it is a buffer solution, so we will apply Henderson HasselBalch equation written below:
 $ pH=p{{K}_{a}}+\log \dfrac{salt}{acid} $
 $ pH=4.2+\log \left( \dfrac{0.001}{0.001} \right) $
 $ \Rightarrow pH=4.2 $
Thus now we have $ pH=7+\dfrac{1}{2}\left[ pKa+\log C \right] $
 $ =7+\dfrac{1}{2}\left[ 4.2+\log 0.01 \right] $
On further simplifying;
 $ =7+\dfrac{1}{2}\left[ \dfrac{4.2}{2} \right] $
 $ =7+\dfrac{1}{2}\left[ 2.2 \right] $
 $ \Rightarrow 8.1 $
Therefore the correct answer is option C.

Note :
In the above problem the mixture is acting as a buffer solution. Buffer solution comprises weak acid and its conjugate base or weak base and its conjugate acid. In the above question sodium acetate is a weak basic salt while acetic acid is a conjugate acid of it. So it acts as a good buffer for weak acidic solutions.