Chapter 14, Problem 76A

Interpretation Introduction

Interpretation:

Volume of stock HNO₃ solution have to be calculated.

Concept introduction:

The equation for dilution is

$\begin{array}{l} M_1{V}_1=M_2{V}_2\hfill \\ Stocksolution=Dilutedsolution\hfill \end{array}$

Concentrated solutions of known molarities are known as stock solution.

Where,

M₁ = molarity of the stock solution V₁ = volume of stock solution

M₂ = molarity of the diluted solution V₂ = volume of diluted solution

Answer to Problem 76A

45 mL of 5.0 M nitric acid is required to make 225 mL 1.0 M HNO₃ solution.

Explanation of Solution

Given information:

Molarity of the stock HNO₃ solution $M_1=5.0M$.

Molarity of the diluted HNO₃ solution $M_2=1.0M$.

Volume of diluted HNO₃ solution $V_2=225mL$.

Write the expression for dilution.

$\begin{array}{l} M_1{V}_1=M_2{V}_2\hfill \\ Stocksolution=Dilutedsolution\hfill \end{array}$

Volume of 5.0 M HNO₃ solution needed to make 1.0 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{1.0M\times 225mL}{5M}=45mL\end{array}$

Thus, 45 mL of 5.0 M nitric acid is required to make 225 mL 1.0 M HNO₃ solution.