Chapter 15, Problem 15.62QA

Interpretation Introduction

To calculate:

a) $\mathrm{p}\mathrm{H}$ of $0.0752\mathrm{M}$ quinoline $\left({\mathrm{C}}_9{\mathrm{H}}_7\mathrm{N}\right)$.

b) The $\left[\mathrm{O}{\mathrm{H}}^{-}\right]$ concentration of solution in part $\mathrm{a}$.

Answer to Problem 15.62QA

Solution:

a) $pH=8.86$

b) $\left[O{H}^{-}\right]=7.30\times {10}^{-6} M$

Explanation of Solution

1) Concept:

We can calculate ${\mathrm{K}}_{\mathrm{b}}$ of quinolone using its $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$. Using the RICE table and ${\mathrm{K}}_{\mathrm{b}}$ expression for reaction, we can calculate the hydroxide ion concentration. From this, we can get $\mathrm{p}\mathrm{O}\mathrm{H}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{p}\mathrm{H}$.

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

Here, ${\mathrm{K}}_{\mathrm{b}}$ is equilibrium constant of base, $\left[\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}\right]$ is the concentration of products, $\left[\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\right]$ is the concentration of reactants, and coefficient are the coefficient of reactants and product from the balanced reaction.

2) Formula:

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

ii) $p{K}_b= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_b\right]$

iii) $pH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[H3O^{+}\right]$

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

v) $pH+pOH=14$

3) Given:

i) $p{K}_{b}of quinoline=9.15$

ii) $\left[C_9{H}_{7}N\right]=0.0752 M$

4) Calculations:

RICE table for the given reaction:

Reaction	$C_9{H}_7{N}_{\left(aq\right)} + H_2{O}_{\left(l\right)} \rightleftharpoons C_9{H}_{7}N{H}_{\left(aq\right)}^{+} + O{H}_{\left(aq\right)}^{-}$
	$\left[C_9{H}_{7}N\right]\left(M\right)$		$\left[C_9{H}_{7}N{H}^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$0.0752$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(0.0752-x)$		$x$	$x$

$p{K}_b= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_b\right]$

$9.15= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_b\right]$

$K_b={10}^{-9.15}=7.08\times {10}^{-10}$

Therefore, $K_b$ expression for this reaction is

$K_b=\frac{\left[C_9{H}_{7}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[C_9{H}_{7}N\right]}=\frac{\left(x\right)\left(x\right)}{(0.0752-x)}$

$7.08\times {10}^{-10}=\frac{x^2}{(0.0752-x)}$

To solve for x, we make the simplifying assumption that x will be small compared to 0.0752 M because ${ \left[C_9{H}_{7}N\right]}_{initial}$ is $0.0752/(7.08\times {10}^{-10}) =1.06\times {10}^8$ times the $K_b$ value, which is much greater than our $500\times$ guideline.

$7.08\times {10}^{-10}=\frac{x^2}{0.0752}$

$7.08\times {10}^{-10}\times 0.0752=x^2$

$x^2=5.32\times {10}^{-11}$

$x=7.30\times {10}^{-6} M$

The ratio of $0.0752/(7.30\times {10}^{-6})$

$\left[O{H}^{-}\right]=x=7.30\times {10}^{-6} M$

$pOH= -\mathit{log}\left[7.30\times {10}^{-6}\right]=5.14$

$pH+5.14=14$

$pH=14-5.14=8.86$

Conclusion:

We got the value of $K_b$ using the given value of $p{K}_b$. Using the RICE table and $K_b$, we got the value $pH$ of solution.