Chapter 20, Problem 105CP

Interpretation Introduction

Interpretation: It is given that, alkaline earth metal reacts with water to produce hydrogen. The mass of unknown alkaline earth metal, volume of water and hydrogen is given. The unknown alkaline earth metal is to be identified. The $\text{pH}$ value of this solution is to be calculated.

Concept introduction: The expression to calculate the number of moles by using the ideal gas law is,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

To determine: The unknown alkaline earth metal that when reacts with water produces hydrogen gas; the $\text{pH}$ value of this solution.

Answer to Problem 105CP

Answer

The unknown earth metal is calcium $\left(\text{Ca}\right)$. The required value of $\text{pH}$ for the given solution is $\underset{\_}{12.698}$.

Explanation of Solution

The number of moles of hydrogen gas is $\underset{\_}{0.249mol}$.

Given

Mass of alkaline earth metal is $10.0\text{g}$.

Volume of water is $10.0\text{L}$.

Volume of hydrogen gas is $6.10\text{L}$.

Pressure is $1\text{atm}$.

Temperature is $25°\text{C}$.

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}$

Hence,

The conversion of $\text{25}°\text{C}$ into Kelvin is,

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}\\ T\left(\text{K}\right)=\left(\text{25}+\text{273}\right)\text{K}\\ =\text{298}\text{K}\end{array}$

Formula

The number of moles of hydrogen gas is calculated using the formula,

$\begin{array}{c}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

Where,

• $P$ is the total pressure.
• $V$ is the volume.
• $n$ is the total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is the absolute temperature.

Substitute the values of $P,V,R$ and $T$ in the above equation.

$\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(\text{1}\text{atm}\right)\left(\text{6}\text{.10}\text{L}\right)}{\left(\text{298}\text{K}\right)\left((\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})\right)}\\ =\underset{\_}{0.249mol}\end{array}$

The unknown earth metal is calcium $\left(\text{Ca}\right)$.

Given

Mass of alkaline earth metal is $10.0\text{g}$.

Volume of water is $10.0\text{L}$.

It is assumed that the unknown metal is $\text{X}$. The reaction of unknown metal with water is,

$\text{X}\left(s\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{H}}_{\text{2}}\left(g\right)+\text{X}{\left(\text{OH}\right)}_{\text{2}}\left(aq\right)$

Since, $0.249\text{mol}$ of hydrogen is formed when metal reacts with water. The number of moles of metal reacted is,

$0.249\text{mol}\text{of}{\text{H}}_2\times \frac{1\text{mol}\text{X}}{1\text{mol}{\text{H}}_2}=0.249\text{mol}\text{of}\text{X}$

Formula

Molar mass of metal is calculated using the formula,

$\text{Molar}\text{mass}\text{of}\text{metal}=\frac{\text{Given}\text{mass }\text{of metal}}{\text{Number}\text{of}\text{moles}\text{of metal}}$

Substitute the values of the given mass and number of moles of $\text{X}$ in the above equation.

$\begin{array}{c}\text{Molar}\text{mass}\text{of}\text{X}=\frac{\text{Given}\text{mass }\text{of X}}{\text{Number}\text{of}\text{moles}\text{of X}}\\ =\frac{10.0\text{g}}{0.249\text{mol}}\\ =\underset{\_}{40.2g/mol}\end{array}$

This molar mass belongs to calcium $\left(\text{Ca}\right)$.

The concentration of ${\text{OH}}^{-}$ is $\underset{\_}{0.0498M}$.

Given

Volume of solution is $10.0\text{L}$.

The number of moles of calcium is $0.249\text{mol}$.

Formula

The concentration of ${\text{Ca(OH)}}_{\text{2}}$ is calculated as,

$\left[{\text{Ca(OH)}}_{\text{2}}\right]=\frac{n}{V}$

Where,

• $n$ is the number of moles of calcium.
• $V$ is the volume of solution.

Substitute the values of $n$ and $V$ in the above equation.

$\begin{array}{c}\left[{\text{Ca(OH)}}_{\text{2}}\right]=\frac{n}{V}\\ =\frac{\text{0}\text{.249}\text{mol}}{\text{10}\text{.0}\text{L}}\\ =0.0249\text{M}\end{array}$

The compound formed ${\text{Ca(OH)}}_{\text{2}}$ is a strong base. Since, two moles of ${\text{OH}}^{-}$ are produced for every one mole of ${\text{Ca(OH)}}_{\text{2}}$, therefore, the concentration of ${\text{OH}}^{-}$ will be twice the concentration of ${\text{Ca(OH)}}_{\text{2}}$, that is,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=2\times 0.0249\text{M}\\ =\underset{\_}{0.0498M}\end{array}$

The value of $\text{pOH}$ is $\underset{\_}{1.302}$.

The concentration of ${\text{OH}}^{-}$ is $0.0498\text{M}$.

Formula

The $\text{pOH}$ is calculated using the formula,

$\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]$

Substitute the value of concentration of ${\text{OH}}^{-}$ in the above equation.

$\begin{array}{c}\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]\\ =-{\log }_{\text{10}}\left[\text{0}\text{.0498}\right]\\ =\underset{\_}{1.302}\end{array}$

The value of $\text{pH}$ for the given solution is $\underset{\_}{12.698}$.

The value of $\text{pOH}$ is $1.302$.

Formula

The relation between $\text{pH}$ and $\text{pOH}$ is,

$\text{pH}+\text{pOH}=\text{14}$

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{pH}+\text{1}\text{.302}=\text{14}\\ \text{pH}=\underset{\_}{12.698}\end{array}$

Conclusion

Conclusion

The molar mass of unknown earth metal is 40.2 g/mol.  This is the molar mass of calcium.  The required $\text{pH}$ value for the given solution is 12.698