Chapter 15, Problem 15.64QA

Interpretation Introduction

To calculate:

a) The $\mathrm{p}\mathrm{H}$$\mathrm{o}\mathrm{f}$$3.00\times {10}^{-4}\mathrm{M}$ trimethylamine $\left[{\left(\mathrm{C}{\mathrm{H}}_3\right)}_3\mathrm{N})\right]\mathrm{i}\mathrm{f}\mathrm{ }\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{ }{\mathrm{K}}_{\mathrm{b}}\mathrm{ }\mathrm{i}\mathrm{s}\mathrm{ }6.5\times {10}^{-5}.$

b) The $\mathrm{p}\mathrm{H}$$\mathrm{o}\mathrm{f}$$2.88\times {10}^{-3}\mathrm{M}$ methylamine $\left[\left(\mathrm{C}{\mathrm{H}}_3\right)\mathrm{N}{\mathrm{H}}_2)\right]\mathrm{i}\mathrm{f}\mathrm{ }\mathrm{i}\mathrm{t}\mathrm{s}\mathrm{ }\mathrm{ }{\mathrm{K}}_{\mathrm{b}}\mathrm{ }\mathrm{i}\mathrm{s}\mathrm{ }4.4\times {10}^{-4}.$

c) The concentration of dimethylamine $\left[{\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{N}\mathrm{H})\right]$ is needed for the solution to have the same $\mathrm{p}\mathrm{H}$ as the solution in part $\mathrm{b}$.

Answer to Problem 15.64QA

Solution:

a) $\mathrm{p}\mathrm{H}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}=10.05$

b) $\mathrm{p}\mathrm{H}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{y}\mathrm{l}\mathrm{a}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{e}=10.97$

c) $\left[{\left(\mathrm{C}{\mathrm{H}}_3\right)}_2\mathrm{N}\mathrm{H})\right]=2.39\times {10}^{-3}\mathrm{ }\mathrm{M}$

Explanation of Solution

1) Concept:

Using RICE table and $K_b$ expression for reaction, we can calculate the hydroxide ion concentration. From this, we can get $pOH and pH$.

2) Formula:

i) $K_b=\frac{{\left[Product\right]}^{coefficents}}{{\left[Reactant\right]}^{coefficents}}$

ii) $pOH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[O{H}^{-}\right]$

iii) $pH+pOH=14$

3) Given:

i) $a) \left[{\left(C{H}_3\right)}_{3}N)\right]=3.00\times {10}^{-4}M, K_b=6.5\times {10}^{-5}$

ii) $b) \left[\left(C{H}_3\right)N{H}_2\right]=2.88\times {10}^{-3}M K_b=4.4\times {10}^{-4}\mathrm{ }$

iii) $c) {K_b of \left(C{H}_3\right)}_{2}NH=5.9\times {10}^{-4}$

4) Calculations:

a) To calculate the pH of trimethylamine, we use RICE table and initial concentrations.

RICE table for the given reaction:

Reaction	${\left(C{H}_3\right)}_3{N}_{\left(aq\right)} + H_2{O}_{\left(l\right)} \rightleftharpoons {\left(C{H}_3\right)}_{3}N{H}_{\left(aq\right)}^{+} + O{H}_{\left(aq\right)}^{-}$
	$\left[{\left(C{H}_3\right)}_{3}N\right]\left(M\right)$		$\left[{\left(C{H}_3\right)}_{3}N{H}^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$3.00\times {10}^{-4}$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(3.00\times {10}^{-4}-x)$		$x$	$x$

$K_b$ expression for this reaction is

$K_b=\frac{\left[{\left(C{H}_3\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_3\right)}_{3}N\right]}=\frac{\left(x\right)\left(x\right)}{(3.00\times {10}^{-4}-x)}$

$6.5\times {10}^{-5}=\frac{x^2}{(3.00\times {10}^{-4}-x)}$

$x^2+6.5\times {10}^{-5}x-1.95\times {10}^{-8}=0$

Solving this quadratic equation, we can get two values of x.

$x=1.11\times {10}^{-4} and x=-1.76\times {10}^{-4}$

Concentration is never negative; therefore, we will use the positive value of $x$.

$\left[O{H}^{-}\right]=x=1.11\times {10}^{-4} M$

$pOH= -\log \left[1.11\times {10}^{-4}\right]=3.95$

$pH+3.95=14$

$pH=14-3.95=10.05$

b) To calculate the pH of methylamine, we will use RICE table and initial concentrations.

RICE table for the given reaction:

Reaction	$\left(C{H}_3\right)N{H}_{2\left(aq\right)} + H_2{O}_{\left(l\right)} \rightleftharpoons \left(C{H}_3\right)N{H}_{3\left(aq\right)}^{+} + O{H}_{\left(aq\right)}^{-}$
	$\left[\left(C{H}_3\right)N{H}_2\right]\left(M\right)$		$\left[\left(C{H}_3\right)N{H}_3^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$2.88\times {10}^{-3}$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(2.88\times {10}^{-3}-x)$		$x$	$x$

$K_b$ expression for this reaction is

$K_b=\frac{\left[\left(C{H}_3\right)N{H}_3^{+}\right]\left[O{H}^{-}\right]}{\left[\left(C{H}_3\right)N{H}_2\right]}=\frac{\left(x\right)\left(x\right)}{(2.88\times {10}^{-3}-x)}$

$4.4\times {10}^{-4}=\frac{x^2}{(2.88\times {10}^{-3}-x)}$

$x^2+4.4\times {10}^{-4}x-1.27\times {10}^{-6}=0$

Solving this quadratic equation, we can get two values of x.

$x=9.28\times {10}^{-4} and x=-1.37\times {10}^{-3}$

Concentration is never negative; therefore, we will use the positive value of $x$.

$\left[O{H}^{-}\right]=x=9.28\times {10}^{-4} M$

$pOH= -\log \left[9.28\times {10}^{-4}\right]=3.03$

$pH+3.03=14$

$pH=14-3.03=10.97$

c) To calculate the concentration of dimethylamine, we use Kb expression and RICE table.

Dimethyl amine solution has the same $pH$ as that of part $b$.

Therefore, $pOH and the \left[O{H}^{-}\right]$ concentration of dimethylamine is also the same as in part $b$.

Let’s assume the initial dimethylamine concentration as $“A”.$

RICE table for given reaction:

Reaction	${\left(C{H}_3\right)}_2{NH}_{\left(aq\right)} + H_2{O}_{\left(l\right)} \rightleftharpoons {\left(C{H}_3\right)}_{2}N{H}_{2\left(aq\right)}^{+} + O{H}_{\left(aq\right)}^{-}$
	$\left[{\left(C{H}_3\right)}_{2}NH\right]\left(M\right)$		$\left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$A$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(A-x)$		$x$	$x=9.28\times {10}^{-4}$

$K_b$ expression for this reaction is

$K_b=\frac{\left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_3\right)}_{2}NH\right]}$

$5.9\times {10}^{-4}=\frac{(9.28\times {{10}^{-4})}^2}{(A-9.28\times {10}^{-4})}$

$5.9\times {10}^{-4}=\frac{8.61\times {10}^{-7}}{(A-9.28\times {10}^{-4})}$

$A-9.28\times {10}^{-4}=\frac{8.61\times {10}^{-7}}{5.9\times {10}^{-4}}$

$A-9.28\times {10}^{-4}=1.46\times {10}^{-3}$

$A=2.39\times {10}^{-3} M$

$\left[{\left(C{H}_3\right)}_{2}NH\right]=2.39\times {10}^{-3} M$

Conclusion:

Using RICE table and the $K_b$ expression, we got the $pH$ of trimethylamine and methylamine. Using the $pH$ and the RICE table, we calculated the concentration of dimethyl amine.