Chapter 15, Problem 15.105QA

Interpretation Introduction

To find:

The pK_a of 0.125(M) aqueous solution of sodium propanoate

Answer to Problem 15.105QA

Solution:

The pK_a for 0.125M aquoues solution of sodium propionate is 8.97

Explanation of Solution

1) Concept:

Sodium propionate is a salt of strong base NaOH and weak acid $C{H}_{3}COOH$.So it is a basic salt.$N{a}^{+}$ ion has no role in its acid-base properties, they are spectator ions.

First, we have to find out the value of $K_a$ from the given value of $p{K}_a$.

Then, molar concentration of $O{H}^{-}$ ion has to be calculated using the RICE table for the hydrolysis of Sodium propanoate salt.

$pOH$ is calculated using the molar concentration of $O{H}^{-}$

And finally, we calculate $pH$ from the $pOH$ value.

2) Formula:

i) $p{K}_a=-\log K_a$

ii) $K_h=\frac{K_w}{K_a}$

iii) $pOH=-\log \left[{OH}^{-}\right]$

iv) $pH= 14.00-pOH$

3) Calculation:

$K_a$ can be calculated as:

$4.85=-\log \left[K_a\right]$

$K_a=1.41\times {10}^{-5}$

$K_h=\frac{K_w}{K_a} =\frac{1.0\times {10}^{-14}}{1.41\times {10}^{-5}}=0.709\times {10}^{-9}$

Now we have to calculate the molar concentration of $C{H}_{3}C{H}_{2}COONa$ using RICE table:

$Reaction: C{H}_3{CH}_2{COO}^{-}\left(aq\right)+H_{2}O\left(l\right) \rightleftharpoons {CH}_{3}C{H}_{2}COOH+{OH}^{-}$

$Initial 0.125 0 0$

$Change -x +x +x$

$Equilibrium (0.125-x) x x$

To solve for ‘x’, a simplifying assumption has to be made that ‘x’ is very small compared to $0.125M$ because ${\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]}_{initial}$ is $\frac{0.125}{0.709 \times 10-9} = 0.176 \times {10}^9$ times than K_b value, which is much greater than 500xguideline.

$K_h=\frac{\left[C{H}_3{CH}_{2}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}C{H}_{2}CO{O}^{-}\right]} = \frac{\left(x\right)\left(x\right)}{\left(0.125-x\right)}\approx \frac{x^2}{0.125}=0.709\times {10}^{-9}$

$x=9.414\times {10}^{-6}$

So, $\left[O{H}^{-}\right]=9.414\times {10}^{-6}\left(M\right)$

Therefore, $pOH=-\mathrm{l}\mathrm{o}\mathrm{g}[9.414\times {10}^{-6}]=6-0.9737= 5.026$

$pH =14.00-5.026=8.97$

Conclusion:

The pH of 0.125(M) aqueous solution of Sodium propanoateis 8.97