Q.10.109 - An Introduction to General, Organic, and Biological Chemistry

Question

A 0.204 M NaOH solution is used to titrate 50.0 mL of an ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution.

a. Write the balanced chemical equation.

b. What is the molarity of the ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution if 16.4mL of the NaOH solution is required?

Step-by-Step Solution

Answer

Part (a). ${\mathrm{H}}_3{\mathrm{PO}}_4+3\mathrm{NaOH}\to 3{\mathrm{H}}_2\mathrm{O}+{\mathrm{Na}}_3{\mathrm{PO}}_4$

Part (b). $0.0022 M$

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 chemical reaction is,

${\mathrm{H}}_3{\mathrm{PO}}_4+3\mathrm{NaOH}\to 3{\mathrm{H}}_2\mathrm{O}+{\mathrm{Na}}_3{\mathrm{PO}}_4$

Step 3: Explanation Part (b)

Given,

Molarity $M_1$=$0.204 M$of NaOH and Volume $V_1=16.4 mL$

Volume $V_2=50 mL$of ${\mathrm{H}}_3{\mathrm{PO}}_4$

Using,

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

Molarity $M_2$of ${\mathrm{H}}_3{\mathrm{PO}}_4$ solution,

$$\begin{gathered} =\frac{16.4 \mathrm{ml}}{50 \mathrm{ml}}\times \frac{0.204 \mathrm{mol}}{1000 \mathrm{ml}}\times \frac{1}{3}\times \frac{1000 \mathrm{ml}}{1 \mathrm{L}} \\ =0.022 M \end{gathered}$$