Question

Given,20 mL of 0.1M ${{H}_{2}}S{{O}_{4}}$ ​solution is added to 30 mL of 0.2M $N{{H}_{4}}OH$ solution. The pH of the resultant mixture is: [ $p{{K}_{b}}$ ($N{{H}_{4}}OH$)​=4.7].
(A) 9.4
(B) 5.0
(C) 9.0
(D) 5.2

Answer 1

Hint: As we know that a buffer solution is an aqueous solution which consists of a mixture of a weak acid and its conjugate base or a mixture of weak base and its conjugate acid. Also the pH of buffer solution changes very little when a minute amount of strong acid or base is added to it.
Formula used:
We will use the following formulas for this question:-
$pH+pOH=14$
$pOH=p{{K}_{b}}+\log \dfrac{\text{ }\!\![\!\!\text{ salt or conjugate acid }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ weak base }\!\!]\!\!\text{ }}$

Complete answer:
Let us understand the concept of buffer solution first:-
-Buffer solution: It is an aqueous solution which consists of a mixture of a weak acid and its conjugate base or a mixture of weak base and its conjugate acid. Also the pH of buffer solution changes very little when a minute amount of strong acid or base is added to it.
-Calculation of millimoles (since the volume id given in milliliters) of reactants:-
Number of moles(n) = $\text{concentration }\!\!\times\!\!\text{ volume }$
(A) ${{H}_{2}}S{{O}_{4}}\to 2{{H}^{+}}+S{{O}_{4}}^{2-}$
Concentration of ${{H}_{2}}S{{O}_{4}}$= 0.1M
Volume of ${{H}_{2}}S{{O}_{4}}$= 20mL
Number of moles of ${{H}_{2}}S{{O}_{4}}$=$\text{concentration }\!\!\times\!\!\text{ volume }$
${{n}_{{{H}_{2}}S{{O}_{4}}}}=0.1M\times 20mL=2\text{millimoles}$
Since we can see, each${{H}_{2}}S{{O}_{4}}$ gives two ions of${{H}^{+}}$. Hence the number of millimoles of ${{H}^{+}}$= 4millimoles.
(B)$N{{H}_{4}}OH\to N{{H}_{4}}^{+}+O{{H}^{-}}$
Concentration of $N{{H}_{4}}OH$= 0.2M
Volume of $N{{H}_{4}}OH$= 30mL
Number of moles of $N{{H}_{4}}OH$=$\text{concentration }\!\!\times\!\!\text{ volume }$
${{n}_{N{{H}_{4}}OH}}=0.2M\times 30mL=6\text{millimoles}$
-Now the reaction between ${{H}_{2}}S{{O}_{4}}$and $N{{H}_{4}}OH$occurs as follows:-
${{H}^{+}}+N{{H}_{4}}OH\rightleftharpoons N{{H}_{4}}^{+}+{{H}_{2}}O$

4	6	-	-
4	6-2 = 2	4	4

From above table we can see, 4 millimoles of reacts ${{H}_{2}}S{{O}_{4}}$with 4 millimoles of$N{{H}_{4}}OH$ to give 4 millimoles of conjugate acid. Hence the solution formed is a basic buffer.
-Now we will use the following formula to calculate the pOH of the solution:-
(Value of $p{{K}_{b}}$ ($N{{H}_{4}}OH$)​=4.7)
\[\begin{align}
  & \Rightarrow pOH=p{{K}_{b}}+\log \dfrac{\text{ }\!\![\!\!\text{ N}{{\text{H}}_{4}}^{+}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ N}{{\text{H}}_{4}}\text{OH }\!\!]\!\!\text{ }} \\
 & \Rightarrow pOH=4.7+\log \dfrac{\text{ }\!\![\!\!\text{ millimoles of N}{{\text{H}}_{4}}^{+}\text{/millivolume }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ millimoles of N}{{\text{H}}_{4}}\text{OH/millivolume }\!\!]\!\!\text{ }} \\
 & \Rightarrow pOH=4.7+\log \dfrac{\text{ }\!\![\!\!\text{ millimoles of N}{{\text{H}}_{4}}^{+}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ millimoles of N}{{\text{H}}_{4}}\text{OH }\!\!]\!\!\text{ }} \\
 & \Rightarrow pOH=4.7+\log \dfrac{4}{2} \\
 & \Rightarrow pOH=4.7+\log 2 \\
 & \Rightarrow pOH=4.7+0.3 \\
 & \Rightarrow pOH=5.0 \\
\end{align}\]
-Calculation of pH of the solution:-
As we know that $pH+pOH=14$
pH = 14 – 5.0 = 9.0

Hence the pH of the resultant mixture is: (C) 9.0

Note:
 -For solving such types of questions, learn and understand the concept of buffer solutions and equilibrium.
-Also remember to use the concentration terms carefully and along with the units so as to get the accurate result.