Question

What is the molarity of an ${H_2}S{O_4}$ solution if $18.5mL$ of $0.18M$ $NaOH$ are needed to neutralize $25mL$ of the sample? ${H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O$ .

Answer 1

Hint: The molar fixation (likewise called molarity, sum focus, or substance focus) is a proportion of the grouping of a synthetic animal categories, specifically of a solute in an answer, as far as the measure of substance per unit volume of arrangement. In science, the most normally utilized unit for molarity is the quantity of moles per liter, having the unit image $mol/L$ or $mol.d{m^{ - 3}}$ in the SI unit. An answer with a centralization of one $mol/L$ is supposed to be one molar, ordinarily assigned as one M.

Complete step by step answer:
We have to see the molarity or molar focus is most normally communicated in units of moles of solute per liter of arrangement. For use in more extensive applications, it is characterized as the measure of substance of solute per unit volume of arrangement, or per unit volume accessible to the species, addressed by lowercase c.
By using the following formula, because the following expression is molarity formula,
$c = \dfrac{n}{v}$
Rewrite the above equation,
$n = cv$
First, we have to calculate the number of moles,
$c = 0.18M$
$v = 18.5mL$
Applying both the values in the above equation,
$nNaOH = \dfrac{{18.5 \times 0.18}}{{1000}}$
The number of moles = $3.33 \times {10^{ - 3}}$,
From the condition, we can see that the quantity of moles ${H_2}S{O_4}$ should be a large portion of that sum,
So that,
$n{H_2}S{O_4} = \dfrac{{3.33}}{2} \times {10^{ - 3}} = 1.66 \times {10^{ - 3}}$
Then, $v = 25mL$
Applying both the values in the molarity equation,
$[{H_2}S{O_4}] = \dfrac{{1.66 \times {{10}^{ - 3}}}}{{25.0 \times {{10}^{ - 3}}}}$
Therefore,
$[{H_2}S{O_4}] = 0.066mol/L$

Note: In thermodynamics, the utilization of molar fixation is frequently not advantageous in light of the fact that the volume of most arrangements somewhat relies upon temperature because of warm extension. This issue is normally settled by presenting temperature adjustment factors, or by utilizing a temperature-free proportion of focus like molality.