Question

Match each product in LIST-I with one or more reaction in LIST-II and choose the correct option.
R is an anion, $p{K}_a,RH=5$

$\begin{array}{cc}\text{Column - I}& \text{Column - II}\\ \text{(1M each)}& \\ \left(P\right)10mLRH+10mLNaOH& \left(1\right)pH<5\\ \left(Q\right)20mLRH+10mLNaOH& \left(2\right)pH=5\\ \left(R\right)10mLRH+20mLNaOH& \left(3\right)pH>5\\ \left(S\right)10mLRH+20mLHCl& \left(4\right)\text{buffer}\end{array}$

• A. $P\to 1;Q\to 2,3;R\to 3,4;S\to 5$
• B. $P\to 3.5;Q\to 2,3;R\to 1,2,3;S\to 4,5$
• C. $P\to 3.5;Q\to 2,3;R\to 1;S\to 5$
• D. $P\to 3.5;Q\to 2,5;R\to 1;S\to 4$

Answer 1

The correct option is D $P\to 3.5;Q\to 2,5;R\to 1;S\to 4$

If a salt is formed, on hydrolysis:
$pH=7-\frac{1}{2}(p{K}_a+logC)$
If a buffer is formed,
$pH=p{K}_a+log\frac{salt}{acid}$

$\left(P\right)10mLRH+10mLNaOH$
Complete neutralisation, salt RNa is formed.
$pH=7-\frac{1}{2}(6+log\frac{1}{2})=4.35$
$\left(Q\right)20mLRH+10mLNaOH$
Salt and acid are left over, equal amounts, buffer is formed.
$pH=5+log1=5$
$\left(R\right)10mLRH+20mLNaOH$
Salt and strong base are left over, equal amounts (10 mmol in 30 mL solution)
The strong base will dominate this mixture, the pH will be greater than 5.
$\left(S\right)10mLRH+20mLHCl$
The strong acid will dominate this mixture, ph <5.