Chapter 3, Problem 3.106PAE

3.106 Twenty-five mL of a 0.388 M solution of Na₂SO₄ is mixed with 35.3 mL of 0.229 M Na₂SO₄. What is the molarity of the resulting solution? Assume that the volumes are additive.

Interpretation Introduction

Interpretation:

Molarity of $N{a}_{2}S{O}_4$ after mixing $25.0mL$ of $0.388M$$N{a}_{2}S{O}_4$ and $35.3mL$ of $0.229M$$N{a}_{2}S{O}_4$ should be determined.

Concept introduction:

• The number of moles of solute per litter of solution is known as molarity of the solution.

• To calculate the mass of solute use the following expression:

  $\text{Amount of solute in g = Number of moles of solutes }\times \text{ molar mass}$

Given:

$25.0mL$ of $0.388M$$N{a}_{2}S{O}_4$ and $35.3mL$ of $0.229M$$N{a}_{2}S{O}_4$

Answer to Problem 3.106PAE

Solution:

The molarity of $N{a}_{2}S{O}_4$ after mixing $25.0mL$ of $0.388M$

$N{a}_{2}S{O}_4$ and $35.3mL$ of $0.229M$$N{a}_{2}S{O}_4$ is $0.295MN{a}_{2}S{O}_4$.

Explanation of Solution

Molarity:

The number of moles of solute per liter of solution is known as molarity of the solution. It symbolized by $M$. Molarity of any solution is a ratio of moles of solutes per liter of solution. It expressed by the following formula:

$\text{Molarity (}M)=\frac{\text{moles of solutes}}{\text{liter of solution}}$

Number of moles of $N{a}_{2}S{O}_4$:

$\begin{array}{l}\text{Number of moles of }N{a}_{2}S{O}_4=\text{molarity }\times \text{ volume in L}\end{array}$

$\begin{array}{l}=0.388MN{a}_{2}S{O}_4\times 25.0mL\times \frac{1.00L}{\text{1000 }mL}\end{array}$

$\begin{array}{l}=0.0097MolesN{a}_{2}S{O}_4\end{array}$

$\begin{array}{l}\text{Number of moles of }N{a}_{2}S{O}_4=\text{molarity }\times \text{ volume in L}\end{array}$

$\begin{array}{l}=0.229MN{a}_{2}S{O}_4\times 35.3mL\times \frac{1.00L}{\text{1000 }mL}\end{array}$

$\begin{array}{l}=0.0081MolesN{a}_{2}S{O}_4\end{array}$

Now calculate the total volume of $N{a}_{2}S{O}_4$as follows:

$25.0mL+35.3mL=60.3mL$

Now calculate the total moles of $N{a}_{2}S{O}_4$as follows:

$0.0097MolesN{a}_{2}S{O}_4+0.0081MolesN{a}_{2}S{O}_4=0.0178MolesN{a}_{2}S{O}_4$

To calculate the new molarity of $N{a}_{2}S{O}_4$uses the following expression:

$\text{Molarity (}M)=\frac{\text{moles of solutes}}{}$

liter of solution

$=\frac{0.0178MolesN{a}_{2}S{O}_4}{60.3mL\times \frac{1.00L}{\text{1000 }mL}}=0.295MN{a}_{2}S{O}_4$

Conclusion

Therefore, the molarity of $N{a}_{2}S{O}_4$ after mixing $25.0mL$ of $0.388M$$N{a}_{2}S{O}_4$ and $35.3mL$ of $0.229M$$N{a}_{2}S{O}_4$ is $0.295MN{a}_{2}S{O}_4$.