Chapter 16, Problem 146E

A student adds $25.0\text{mL}$ of $0.350\text{M}$ sodium hydroxide to $45.0\text{mL}$ of $0.125\text{M}$ copper $\left(\text{II}\right)$ sulfate. How many grams of copper $\left(\text{II}\right)$ hydroxide will precipitate?

Interpretation Introduction

Interpretation:

The mass of copper $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25\text{mL}$ of $0.350\text{M}$ sodium hydroxide with $45.0\text{mL}$ of $0.125\text{M}$ copper $\left(\text{II}\right)$ sulfate is to be calculated.

Concept introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{n}=\frac{\text{m}}{\text{Molar}\text{mass}}$

Answer to Problem 146E

The mass of copper $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25\text{mL}$ of $0.350\text{M}$ sodium hydroxide and $45.0\text{mL}$ of $0.125\text{M}$ copper $\left(\text{II}\right)$ sulfate is $0.427\text{g}$.

Explanation of Solution

The molarity sodium hydroxide solution is $0.350\text{M}$.

The volume of sodium hydroxide solution is $25\text{mL}$.

The conversion of volume in $\text{L}$ is shown below.

$\begin{array}{rll}\text{V}& =& \left(25\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 25\times {10}^{-3}\text{L}\end{array}$

The molarity copper $\left(\text{II}\right)$ sulfate solution is $0.125\text{M}$.

The volume of copper $\left(\text{II}\right)$ sulfate solution is $45.0\text{mL}$.

The conversion of volume in $\text{L}$ is shown below.

$\begin{array}{rll}\text{V}& =& \left(45.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 45.0\times {10}^{-3}\text{L}\end{array}$

The molar mass of copper $\left(\text{II}\right)$ hydroxide is $97.57\text{g}/\text{mol}$.

The molarity of a solution is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ is the number of moles of the solute.

• $\text{V}$ is the volume of the solution.

Rearrange the above equation for the value of $\text{n}$.

$\text{n}=\text{MV}$

Substitute the values of molarity and volume of sodium hydroxide solution in above expression.

$\begin{array}{rll}\text{n}& =& \left(0.350\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(25\times {10}^{-3}\text{L}\right)\\ & =& 8.75\times {10}^{-3}\text{mol}\end{array}$

The number of moles of sodium hydroxide present in solution is $8.75\times {10}^{-3}\text{mol}$.

Substitute the values of molarity and volume of copper $\left(\text{II}\right)$ sulfate solution in above expression.

$\begin{array}{rll}\text{n}& =& \left(0.125\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(45.0\times {10}^{-3}\text{L}\right)\\ & =& 5.625\times {10}^{-3}\text{mol}\end{array}$

The number of moles of copper $\left(\text{II}\right)$ sulfate present in solution is $5.625\times {10}^{-3}\text{mol}$.

The reaction between sodium hydroxide and copper $\left(\text{II}\right)$ sulfate is shown below.

$2\text{NaOH}+{\text{CuSO}}_{\text{4}}\to \text{Cu}{\left(\text{OH}\right)}_{\text{2}}+{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}$

Two moles of sodium hydroxide reacts with one mole of copper $\left(\text{II}\right)$ sulfate. The available number of moles of copper $\left(\text{II}\right)$ sulfate is more than half of number of moles of sodium hydroxide. Therefore, sodium hydroxide is the limiting reagent of the reaction.

Two moles of sodium hydroxide produced one mole of copper $\left(\text{II}\right)$ hydroxide. Therefore, the relation between the number of moles of sodium hydroxide and copper $\left(\text{II}\right)$ hydroxide is given by the expression as shown below.

${\text{n}}_{\text{Cu}{\left(\text{OH}\right)}_2}=\frac{{\text{n}}_{\text{NaOH}}}{2}$…(1)

Where,

• ${\text{n}}_{\text{NaOH}}$ is the number of moles of sodium hydroxide.

• ${\text{n}}_{\text{Cu}{\left(\text{OH}\right)}_2}$ is the number of moles of copper $\left(\text{II}\right)$ hydroxide.

Substitute the value of ${\text{n}}_{\text{NaOH}}$ in the equation (1).

$\begin{array}{rll}{\text{n}}_{\text{Cu}{\left(\text{OH}\right)}_2}& =& \frac{8.75\times {10}^{-3}\text{mol}}{2}\\ & =& 4.375\times {10}^{-3}\text{mol}\end{array}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{m}=\text{n}\times \text{Molar}\text{mass}$

Where,

• $\text{m}$ is the mass of the substance.

• $\text{n}$ is the number of moles of the substance.

Substitute the value of number of moles and molar mass of copper $\left(\text{II}\right)$ hydroxide in the above equation.

$\begin{array}{rll}\text{m}& =& \left(4.375\times {10}^{-3}\text{mol}\right)\left(97.57\text{g}/\text{mol}\right)\\ & =& 0.427\text{g}\end{array}$

Therefore, the mass of copper $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25\text{mL}$ of $0.350\text{M}$ sodium hydroxide and $45.0\text{mL}$ of $0.125\text{M}$ copper $\left(\text{II}\right)$ sulfate is $0.427\text{g}$.

Conclusion

The mass of copper $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25\text{mL}$ of $0.350\text{M}$ sodium hydroxide and $45.0\text{mL}$ of $0.125\text{M}$ copper $\left(\text{II}\right)$ sulfate is $0.427\text{g}$.