Chapter 16, Problem 147E

Interpretation Introduction

Interpretation:

The mass of nickel $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25.0\text{mL}$ of $0.269\text{M}$ nickel $\left(\text{II}\right)$ chloride with $30.0\text{mL}$ of $0.260\text{M}$ potassium hydroxide 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 147E

The mass of nickel $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25.0\text{mL}$ of $0.269\text{M}$ nickel $\left(\text{II}\right)$ chloride and $30.0\text{mL}$ of $0.260\text{M}$ potassium hydroxide is $0.36\text{g}$.

Explanation of Solution

The molarity potassium hydroxide solution is $0.260\text{M}$.

The volume of potassium hydroxide solution is $30.0\text{mL}$.

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

$\begin{array}{rll}\text{V}& =& \left(30.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 30.0\times {10}^{-3}\text{L}\end{array}$

The molarity nickel $\left(\text{II}\right)$ chloride solution is $0.269\text{M}$.

The volume of nickel $\left(\text{II}\right)$ chloride solution is $25.0\text{mL}$.

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

$\begin{array}{rll}\text{V}& =& \left(25.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 25.0\times {10}^{-3}\text{L}\end{array}$

The molar mass of nickel $\left(\text{II}\right)$ hydroxide is $92.708\text{g}/\text{mol}$.

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 potassium hydroxide solution in above expression.

$\begin{array}{rll}\text{n}& =& \left(0.260\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(30.0\times {10}^{-3}\text{L}\right)\\ & =& 7.8\times {10}^{-3}\text{mol}\end{array}$

The number of moles of potassium hydroxide present in solution is $7.8\times {10}^{-3}\text{mol}$.

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

$\begin{array}{rll}\text{n}& =& \left(0.269\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(25.0\times {10}^{-3}\text{L}\right)\\ & =& 6.725\times {10}^{-3}\text{mol}\end{array}$

The number of moles of nickel $\left(\text{II}\right)$ chloride present in solution is $6.725\times {10}^{-3}\text{mol}$.

The reaction between potassium hydroxide and nickel $\left(\text{II}\right)$ chloride is shown below.

$2\text{KOH}+{\text{NiCl}}_2\to \text{Ni}{\left(\text{OH}\right)}_{\text{2}}+2\text{KCl}$

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

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

${\text{n}}_{\text{Ni}{\left(\text{OH}\right)}_{\text{2}}}=\frac{{\text{n}}_{\text{KOH}}}{2}$…(1)

Where,

• ${\text{n}}_{\text{KOH}}$ is the number of moles of potassium hydroxide.

• ${\text{n}}_{\text{Ni}{\left(\text{OH}\right)}_{\text{2}}}$ is the number of moles of nickel $\left(\text{II}\right)$ hydroxide.

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

$\begin{array}{rll}{\text{n}}_{\text{Ni}{\left(\text{OH}\right)}_{\text{2}}}& =& \frac{7.8\times {10}^{-3}\text{mol}}{2}\\ & =& 3.9\times {10}^{-3}\text{mol}\end{array}$

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 nickel $\left(\text{II}\right)$ hydroxide in the above equation.

$\begin{array}{rll}\text{m}& =& \left(3.9\times {10}^{-3}\text{mol}\right)\left(92.708\text{g}/\text{mol}\right)\\ & =& 0.36\text{g}\end{array}$

Therefore, the mass of nickel $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25.0\text{mL}$ of $0.269\text{M}$ nickel $\left(\text{II}\right)$ chloride and $30.0\text{mL}$ of $0.260\text{M}$ potassium hydroxide is $0.36\text{g}$.

Conclusion

The mass of nickel $\left(\text{II}\right)$ hydroxide that would be precipitate by the reaction of $25.0\text{mL}$ of $0.269\text{M}$ nickel $\left(\text{II}\right)$ chloride and $30.0\text{mL}$ of $0.260\text{M}$ potassium hydroxide is $0.36\text{g}$.