Chapter 11, Problem 76QP

Interpretation Introduction

Interpretation:

The volume of ${\text{H}}_2{\text{SO}}_4$ required to neutralize given $\text{NaOH}$ solution is to be calculated.

Concept Introduction:

$\text{NaOH}$ is a strong base and ${\text{H}}_2{\text{SO}}_4$ is a strong acid. They both react with each other to produce salt and water.

Answer to Problem 76QP

Solution:

$1.21\text{mL}$ of $0.850{\text{M H}}_2{\text{SO}}_4$ is required to neutralize $20\text{mL}$ of $0.1033\text{M}\text{NaOH}$ solution.

Explanation of Solution

Given Information: $20.00\text{mL}$ of $0.1033\text{M}\text{NaOH}$ is neutralized with $0.850{\text{M H}}_2{\text{SO}}_4$ solution.

:

The balanced chemical equation is,

${\text{H}}_2{\text{SO}}_4(aq)+2\text{NaOH}(aq)\to {\text{Na}}_2{\text{SO}}_4(aq)+2{\text{H}}_2\text{O}(l)$

The molarity of $\text{NaOH}$ solution is $0.1033\text{M}$ . This means that $0.1033\text{mol}$ of $\text{NaOH}$ is present per liter in the solution. The volume of $0.1033\text{M}$ solution is $20\text{mL}$ , so this must be converted into liters. The relationship between molarity, moles of solute, and volume of solution is,

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution in liter}}\mathrm{......}\left(1\right)$

So, moles of solute can be calculated by rearranging equation $\left(1\right)$ .

$\text{Moles of solute}=\text{Molarity}\times \text{Volume of solution in liter }\mathrm{......}\left(2\right)$

Moles of $0.1033\text{M}\text{NaOH}$ present in the $20\text{mL}$ solution are calculated using equation $\left(2\right)$ .

$\begin{array}{c}\text{Moles of NaOH}=20.0\text{mL}\times \frac{1\text{L}}{1000\text{mL}}\times \frac{0.1033\text{mol}}{1\text{L}}\\ =0.002066\text{mol}\end{array}$

It is apparent from the balanced equation that $1\text{mol}$ of ${\text{H}}_2{\text{SO}}_4$ reacts with $2\text{mol}$ $\text{NaOH}$ . So, the moles of ${\text{H}}_2{\text{SO}}_4$ is,

$\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_4=0.002066\text{mol NaOH}\times \frac{1{\text{mol H}}_2{\text{SO}}_4}{2\text{mol NaOH}}\\ =0.001033\text{mol}\end{array}$

Equation $\left(1\right)$ can also be rearranged to the following equation.

$\text{Volume in liter}=\frac{\text{Moles of solute}}{\text{Molarity}}\mathrm{......}\left(3\right)$

Moles of ${\text{H}}_2{\text{SO}}_4$ is $0.001033\text{mol}$ and molarity is $0.850\text{M}$ . So, we put all the values in equation $\left(3\right)$ to get the volume of ${\text{H}}_2{\text{SO}}_4$ :

$\begin{array}{c}{\text{Volume of H}}_2{\text{SO}}_4\text{ in litre}=\frac{0.001033\text{mol}}{\text{0}\text{.850}{\text{mol L}}^{-1}}\\ =0.00121\text{L}\end{array}$

Hence, $0.00121\text{L}$ of $0.850\text{M}$ ${\text{H}}_2{\text{SO}}_4$ are required to neutralize $20\text{mL}\text{of}$ the $\text{NaOH}$ solution. Since the volume in liter is very small, it is converted to milliliter.

$\begin{array}{c}{\text{Volume of H}}_2{\text{SO}}_4\text{ in milliliter}=00.00121\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =1.21\text{mL}\end{array}$

Conclusion

$1.21\text{mL}$ of $0.850{\text{M H}}_2{\text{SO}}_4$ is required to neutralize $0.1033\text{M}\text{NaOH}$ solution.