Chapter 14.2, Problem 25PP

Interpretation Introduction

Interpretation:

Volume of 5.0 M H₂SO₄ stock solutionis to be calculated to prepare 100 mL of 0.25 M H₂SO₄ solution.

Concept introduction:

In dilution, the amount of solute does not change, the number of moles is the same before and after dilution.

If subscript "1" represents initial and "2" represents the final values of the quantities involved, we have:

${\mathrm{n}}_1 =M_1\times V_1 and n_2=M_2\times V_2$

Here, molarity is used as a unit of concentration.

${\mathrm{n}}_1 =n_2$    and thus,

The equation for dilution is

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ Stocksolution=Dilutedsolution\end{array}$

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 25PP

5 mL of 5.0 M H₂SO₄ stock solutionis required to prepare 100 mL of 0.25 M H₂SO₄ solution.

Given information:

The molarity of the stock solution M₁ = 5.0 M

The molarity of the diluted solution M₂ = 0.25 M

The volume of diluted solution V₂ = 100 mL

Explanation:

The equation for dilution is

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ Stocksolution=Dilutedsolution\end{array}$

Where

M₁ = molarity of the stock solution

V₁ = volume of stock solution

M₂ = molarity of the diluted solution

V₂ = volume of diluted solution

V₁ = volume of stock ${\mathrm{H}}_{2}S{O}_4$ solution is to be calculated

${\mathrm{V}}_1=\frac{Molarityofdilutesolution(M_2)}{Molarityofstocksolution(M_1)}\times Volumeofdilutesolution(V_2)$

$\begin{array}{l}=\frac{0.25M\times 100mL}{5M}\\ =5mL\end{array}$

Thus, 5 mL of 5.0 M H₂SO₄ stock solutionis required to prepare 100 mL of 0.25 M ${\mathrm{H}}_{2}S{O}_4$ solution.

Explanation of Solution

Given information:

The molarity of the stock solution M₁ = 5.0 M

The molarity of the diluted solution M₂ = 0.25 M

The volume of diluted solution V₂ = 100 mL

The equation for dilution is

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ Stocksolution=Dilutedsolution\end{array}$

Where

M₁ = molarity of the stock solution

V₁ = volume of stock solution

M₂ = molarity of the diluted solution

V₂ = volume of diluted solution

V₁ = volume of stock ${\mathrm{H}}_{2}S{O}_4$ solution is to be calculated

${\mathrm{V}}_1=\frac{Molarityofdilutesolution(M_2)}{Molarityofstocksolution(M_1)}\times Volumeofdilutesolution(V_2)$

$\begin{array}{l}=\frac{0.25M\times 100mL}{5M}\\ =5mL\end{array}$

Thus, 5 mL of 5.0 M H₂SO₄ stock solutionis required to prepare 100 mL of 0.25 M ${\mathrm{H}}_{2}S{O}_4$ solution.