Chapter 14.2, Problem 24PP

Interpretation Introduction

Interpretation:

Volume of 3.0 M KI stock solutionis to be calculated to prepare 0.3 L of 1.25 M KIsolution.

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:

$n_1 =M_1 \times V_1 and n_2=M_2 \times V_2$

Here, molarity is used as a unit of concentration.

$n_1 =n_2$ and thus,

The equation for dilution is M₁V₁ = M₂V₂

Stock solution = Diluted 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 24PP

125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.

Given information:

The molarity of the stock solution M₁ = 3.0 M The molarity of the diluted solution M₂ = 1.25 M The volume of diluted solution V₂ = 0.3 L

= 300 mL

Explanation:

The equation for dilution is as follows:

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

Here,

M₁ = molarity of the stock solution,

V₁ = volume of stock solution,

M₂ = molarity of the diluted solution,

V₂ = volume of diluted solution.

Volume of stock KI solution =

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

$\begin{array}{l}=\frac{1.25M\times 300mL}{0.3M}\\ =125mL\end{array}$

Thus, 125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.

Explanation of Solution

Given information:

The molarity of the stock solution M₁ = 3.0 M The molarity of the diluted solution M₂ = 1.25 M The volume of diluted solution V₂ = 0.3 L

= 300 mL

The equation for dilution is as follows:

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

Here,

M₁ = molarity of the stock solution,

V₁ = volume of stock solution,

M₂ = molarity of the diluted solution,

V₂ = volume of diluted solution.

Volume of stock KI solution =

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

$\begin{array}{l}=\frac{1.25M\times 300mL}{0.3M}\\ =125mL\end{array}$

Thus, 125 mL of 3.0 M KI stock solutionis required to prepare 0.3 L of 1.25 M KIsolution.