Question

What volume of \[\text{0}\text{.10 M }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]must be added to 50 mL of a 0.10 M NaOH solution to make a solution in which the molarity of the \[\text{ }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\]​ is 0.050 M?
(A) 400 ml
(B) 50 ml
(C) 100 ml
(D) 150 ml

Answer 1

Hint: Molar concentration is a measurement of a chemical species' concentration in a solution, specifically a solute's concentration, in terms of the amount of substance per unit volume of solution. The number of moles per litre, abbreviated as mol/L in SI units, is the most widely used unit for molarity in chemistry. We use this concept here.

Complete Step By Step Answer:
Molar concentration, also known as molarity, is measured in moles of solute per litre of solution. It is defined as the quantity of solute material per unit volume of solution, or per unit volume accessible to the species, for use in larger applications. The use of molar concentration in thermodynamics is frequently inconvenient since the volume of most solutions varies somewhat with temperature owing to thermal expansion. This difficulty is typically handled by utilising temperature adjustment factors or a temperature-independent concentration measure such as molality.
If we add $\mathrm{V} \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ then we are adding 0.01 moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$.
in $50 \mathrm{~mL}$ of 0.1 M $\mathrm{NaOH}$, there are 0.005 moles of $\mathrm{NaOH}$.
The chemical equation for that is :
$2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O} .$
0.0025 moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will neutralize 0.005 moles of $\mathrm{NaOH}$.
So from 0.01 V moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ added in V mL there will be
$0.01 \mathrm{~V}-0.0025$ moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ left unused in the total volume of $\mathrm{V}+50 \mathrm{~mL}$ of solutions.
The conc. there will be $\dfrac{0.01 \mathrm{~V}-0.0025}{\mathrm{~V}+50} \times 1000=0.05 \mathrm{M}$.
$\dfrac{100 \mathrm{v}-2.5}{\mathrm{~V}+50}=0.05$
On solving for $\mathrm{V}$, we get the value:
$\mathrm{V}=0.05002 \mathrm{~L}$
$\mathrm{V}=50.02 \mathrm{~mL}$
Hence, the correct option is $\mathrm{B}$.

Note:
The total molar concentration is calculated by multiplying the density of the mixture by the molar mass of the mixture, also known as the reciprocal of the molar volume of the mixture. The sum of the molar concentrations of salts determines the ionic strength of an ionic solution.