Question

If a 20ml of 0.1 M ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution is added to 30ml of 0.2M ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ solution. What will be the pH of the resultant mixture? [${\text{p}}{{\text{k}}_{\text{b}}}$ of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$= 4.7]
(A) 5.0
(B) 5.2
(C) 9.4
(D) 9.0

Answer 1

pH of a solution can be calculated using the pOH value of the solution, pH of a solution indicates the acidic and alkaline behavior of the solution, a pH value greater than 7 means the solution is alkaline and pH value less than 7 means the solution is acidic.

Formula used: 1. pOH = ${\text{p}}{{\text{k}}_{\text{b}}}$ + $\log \dfrac{{[{{\text{B}}^ + }]}}{{[{\text{BOH}}]}}$ , where [${{\text{B}}^ + }$] = Concentration of conjugate acid
[BOH] = concentration of base
${\text{p}}{{\text{k}}_{\text{b}}}$ = base dissociation constant
2. pH + pOH = 14

Complete step-by-step solution: The mixture of acid ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ and base ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ will result in the formation of a salt, the reaction involved can be written as;

${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + \,\;{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}} \to {{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{ }}{{\text{H}}_{\text{2}}}{\text{O}}$

Now, we can use the formula of molarity to calculate the number of moles of ${{\text{H}}^ + }$ions and ${\text{O}}{{\text{H}}^ - }$ions;

$\Rightarrow {\text{Molarity = }}\frac{{{\text{number of moles }}}}{{{\text{ volume in litre}}}}$

$\Rightarrow$No. of mole = Molarity $\times$ volume in L (since the volume is given in millilitre in the question, therefore we will get the millimoles value)

Number of millimoles of ${{\text{H}}^ + }$ in ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$

1 mole ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will dissociate to give 2 moles of${{\text{H}}^ + }$, ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 2{{\text{H}}^ + }$, hence we multiply 2 with the equation

$\Rightarrow$No. of millimoles ${{\text{H}}^ + }$ = $2 \times 20 \times 0.1$

$\Rightarrow$No. of millimoles${{\text{H}}^ + }$= 4

And 1 mol of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$will dissociate to give 1 mole of${\text{O}}{{\text{H}}^ - }$, ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}} \to {\text{O}}{{\text{H}}^ - }$, hence we multiple 1 with the equation.

Number of millimoles of ${\text{O}}{{\text{H}}^ - }$ in ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$

$\Rightarrow$ No. of millimoles ${\text{O}}{{\text{H}}^ - }$= $1 \times 30 \times 0.2$

$\Rightarrow$No. of millimoles ${\text{O}}{{\text{H}}^ - }$ = 6

The 4 moles of ${{\text{H}}^ + }$ and ${\text{O}}{{\text{H}}^ - }$ will get consumed to form 4 moles of ${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ salt

Therefore the 2 moles of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ base will be remaining in the solution and we will calculate the pOH value from this knowledge,

$\Rightarrow$pOH = ${\text{p}}{{\text{k}}_{\text{b}}}$ + $\log \frac{{[{{{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}]}}{{[{\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}]}}$ (a conjugate acid has more number of oxygen atoms than the parent base)

$\Rightarrow$ pOH = 4.7 + $\log \frac{{[4]}}{{[2]}}$

$\Rightarrow$ pOH = 4.7 + $\log 2$

$\Rightarrow$ pOH = 4.7 + 0.3

$\Rightarrow$ pOH = 5

The pOH value for the solution is 5, but we need the pH value, therefore we will use the 2nd formula mention above

$\Rightarrow$ pH + pOH = 14

$\Rightarrow$ pH = 14 – pOH

$\Rightarrow$pH = 14 – 5

$\Rightarrow$pH = 9

Hence, the correct answer is option (D) i.e.9

Note: Since the pOH value is also mentioned in the given options, this could rise to the selection of the wrong option. Hence, be careful while solving and look for what question is asking, if the question had asked to get the pOH value then we would have stopped our calculation then and there.