Chapter 16, Problem 161AP

Interpretation Introduction

Interpretation:

The mass of ${\text{CH}}_3{\text{NH}}_2$ present in a $100.0\text{ mL}$ solution, which has the pH value of $10.64$, is to be determined.

Concept introduction:

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in the solution from the pH is:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$ as $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\left[{\text{H}}^{+}\right]$

$\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\text{pH}}$…… (1)

The relationship between $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ is:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$…… (2)

The value of ${\text{K}}_{\text{w}}$ is $1.0\times {10}^{-14}$.

The ionization of the weak base takes place as:

$\text{B}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

$K_b$ is the measure of ionization of a base and is known as base-ionization constant, which is specific at a particular temperature:

$K_b=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$…… (3)

The number of moles $\left(n\right)$ of any substance from its given mass $\left(m\right)$ and molar mass $\left(\text{M}\right)$ is calculated as:

$n=\frac{m}{\text{M}}$…… (4)

Here, $m$ is the given mass and $\text{M}$ is the molar mass.

Molarity $\left(M\right)$ of a solution is calculated as:

$M=\frac{n}{V}$…… (5)

Here, $n$ is the number of moles and $V$ is the volume.

Answer to Problem 161AP

Solution: $2.7\times {10}^{-3}\text{ g}$.

Explanation of Solution

Given information:

The pH of methylamine ${\text{CH}}_3{\text{NH}}_2$ solution is $10.64$. Also, the volume of the solution is given as $100.0\text{ mL}$.

Use equation (1) to calculate the concentration of hydronium ions in the solution with the use of pH as:

$\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\text{pH}}$

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-10.64}\\ =2.29\times {10}^{-11}\text{M}\end{array}$

Now, use equation (2) to calculate the value of $\left[{\text{OH}}^{-}\right]$, and substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ and $K_{\text{w}}$ as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

$\begin{array}{c}1.0\times {10}^{-14}=\left(2.29\times {10}^{-11}\right)\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{2.29\times {10}^{-11}}\\ \left[{\text{OH}}^{-}\right]=4.4\times {10}^{-4}\text{M}\end{array}$

${\text{CH}}_3{\text{NH}}_2$ is a weak base and when a weak base is dissolved in water, only partial ionization takes place into ${\text{OH}}^{-}$ ions and its conjugate acid. Thus, the concentration of ${\text{CH}}_3{\text{NH}}_2$ is determined by $K_b$ of the base.

Refer to table $16.7$ for the $K_b$ value as $4.4\times {10}^{-4}$.

The reaction of methylamine is depicted as:

${\text{CH}}_3{\text{NH}}_2\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_3{\text{NH}}_3{}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Prepare an equilibrium table and represent each of the species in terms of $x$ as shown below.

$\begin{array}{ccccc}& {\text{CH}}_3{\text{NH}}_2\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{CH}}_3{\text{NH}}_3{}^{\text{+}}\left(aq\right)& {\text{OH}}^{-}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& x& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -4.4\times {10}^{-4}& -& +4.4\times {10}^{-4}& +4.4\times {10}^{-4}\\ \text{Equilibrium concentration}\left(\text{M}\right)& x-\left(4.4\times {10}^{-4}\right)& -& 4.4\times {10}^{-4}& 4.4\times {10}^{-4}\end{array}$

Now, substitute these concentrations in equation (3) as follows:

$K_b=\frac{\left[{\text{BH}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

$K_b=\frac{\left(4.4\times {10}^{-4}\right)\left(4.4\times {10}^{-4}\right)}{\left(x-\left(4.4\times {10}^{-4}\right)\right)}$

Now, substitute the value of $K_b$ in the above equation as follows:

$\begin{array}{c}4.4\times {10}^{-4}=\frac{\left(4.4\times {10}^{-4}\right)\left(4.4\times {10}^{-4}\right)}{\left(x-\left(4.4\times {10}^{-4}\right)\right)}\\ x-\left(4.4\times {10}^{-4}\right)=\frac{\left(\cancel{4.4\times {10}^{-4}}\right)\left(4.4\times {10}^{-4}\right)}{\left(\cancel{4.4\times {10}^{-4}}\right)}\\ x=8.8\times {10}^{-4}\end{array}$

Thus,

$\left[{\text{CH}}_3{\text{NH}}_2\right]=8.8\times {10}^{-4}\text{ M}$.

The molarity of methylamine is $8.8\times {10}^{-4}\text{ M}$. Now, use equation (5) to calculate the number of moles of ${\text{CH}}_3{\text{NH}}_2$ for $100.0\text{ mL}$:

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

Here, $n$ is the number of moles and $V$ is the volume.

$\begin{array}{c}8.8\times {10}^{-4}\text{ M}=\frac{n}{0.100\text{ L}}\\ n=\left(8.8\times {10}^{-4}\text{ M}\right)\left(0.100\text{ L}\right)\\ n=8.8\times {10}^{-5}\text{ mol}\end{array}$

The molar mass of ${\text{CH}}_3{\text{NH}}_2$ is $31.06\text{ g/mol}$. Now, use equation (4) to calculate the mass of ${\text{CH}}_3{\text{NH}}_2$ as follows:

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

Here, $m$ is the given mass and $\text{M}$ is the molar mass.

$\begin{array}{c}8.8\times {10}^{-5}\text{ mol}=\frac{m}{31.06\text{ g/mol}}\\ m=\left(8.8\times {10}^{-5}\cancel{\text{mol}}\right)\left(31.06\text{ g/}\cancel{\text{mol}}\right)\\ m=2.7\times {10}^{-3}\text{ g}\end{array}$

Conclusion

The mass for the methylamine solution in water is $2.7\times {10}^{-3}\text{ g}$.