Chapter 14, Problem 75A

Stock solutions of HCl with various molarities are frequentlyprepared. Complete Table 14.7 by calculatingthe volume of concentrated, or 12M, hydrochloric acidthat should be used to make 1.0 L of HCl solution witheach molarity listed.

Interpretation Introduction

Interpretation:

Volume of the stocksolutions 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 75A

Molarity of HCl desired	Volume of 12M HCl stock solution needed (mL)
0.5	41.7
1.0	83.3
1.5	125
2.0	167
5.0	417

Explanation of Solution

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

Data given:

M₁ = molarity of the stock solution = 12 M

M₂ = molarity of the diluted solution

V₂ = volume of diluted solution = 1 L

1. Volume of 12M HCl stock solution needed to make 0.5 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{0.5M\times 1L}{12M}=0.0417L=41.7mL\end{array}$

2. Volume of 12M HCl stock solution needed to make 1.0 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{1.0M\times 1L}{12M}=0.0833L=83.3mL\end{array}$

3. Volume of 12M HCl stock solution needed to make 1.5 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{1.5M\times 1L}{12M}=0.125L=125mL\end{array}$

4. Volume of 12M HCl stock solution needed to make 2.0 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{2.0M\times 1L}{12M}=0.167L=167mL\end{array}$

5. Volume of 12M HCl stock solution needed to make 5.0 M diluted solution

$\begin{array}{l}{V}_1=\frac{M_2{V}_2}{M_1}\\ V_1=\frac{5.0M\times 1L}{12M}=0.417L=417mL\end{array}$