Q.10.110 - An Introduction to General, Organic, and Biological Chemistry

Question

A 0.312 M KOH solution is used to titrate 15 mL of an ${\mathrm{H}}_2{\mathrm{SO}}_4$ solution.

a. Write the balanced chemical equation.

b. What is the molarity of the ${\mathrm{H}}_2{\mathrm{SO}}_4$solution if 28.2 mL of KOH solution is required?

Step-by-Step Solution

Answer

Part (a). ${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{KOH}\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{K}}_2{\mathrm{SO}}_4$

Part (b). $0.29 M$

Step 1: Given Information (part a)

A chemical reaction is an interaction wherein at least one substance, likewise called reactants, is switched over completely to at least one distinct substance, known as items.

Step 2: Explanation Part (a)

The balanced equation is,

${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{KOH}\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{K}}_2{\mathrm{SO}}_4$

Step 3: Explanation Part (b)

Given,

Molarity $M_1$=$0.312$KOH solution and Volume$V_1=15 mL$

Volume$V_2=28.2 mL$

Using,

$M_1{V}_1=M_2{V}_2$

molarity of the $H_{2}S{O}_4$ solution 

$$\begin{gathered} =\frac{28.2 \mathrm{ml}}{15 \mathrm{ml}}\times \frac{0.312 \mathrm{mol}}{1000 \mathrm{ml}}\times \frac{1}{2}\times \frac{1000 \mathrm{ml}}{1 \mathrm{L}} \\ =0.29 \mathrm{M} \end{gathered}$$