Answer
(a) 10.3%
(b) 41.58%
(c) 0.986%
Work Step by Step
Ka (formic acid) = $1.7 \times 10^{-4}$
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is at the end of this answer.
-$[H_3O^+] = [HCOO^-] = x$
-$[HCOOH] = [HCOOH]_{initial} - x$
For approximation, we consider: $[HCOOH] = [HCOOH]_{initial}$
(a)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$
$Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 1.6 \times 10^{- 2}}$
$Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 1.6 \times 10^{- 2}}$
$ 2.72 \times 10^{- 6} = x^2$
$x = 1.64 \times 10^{- 3}$
%Ionization: $\frac{ 1.64 \times 10^{- 3}}{ 1.6 \times 10^{- 2}} \times 100\% = 10.3\%$
(b)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$
$Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 5.7 \times 10^{- 4}}$
$Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 5.7 \times 10^{- 4}}$
$ 9.69 \times 10^{- 8} = x^2$
$x = 3.11 \times 10^{- 4}$
5% test: $\frac{ 3.11 \times 10^{- 4}}{ 5.7 \times 10^{- 4}} \times 100\% = 54.6\%$
High percent ionization : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 5.7 \times 10^{- 4}- x}$
$ 9.69 \times 10^{- 8} - 1.7 \times 10^{- 4}x = x^2$
$ 9.69 \times 10^{- 8} - 1.7 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 1.7 \times 10^{- 4})^2 - 4 * (-1) *( 9.69 \times 10^{- 8})$
$\Delta = 2.89 \times 10^{- 8} + 3.87 \times 10^{- 7} = 4.16 \times 10^{- 7}$
$x_1 = \frac{ - (- 1.7 \times 10^{- 4})+ \sqrt { 4.16 \times 10^{- 7}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 1.7 \times 10^{- 4})- \sqrt { 4.16 \times 10^{- 7}}}{2*(-1)}$
$x_1 = - 4.07 \times 10^{- 4} (Negative)$
$x_2 = 2.37 \times 10^{- 4}$
Percent Ionization: $\frac{2.37 \times 10^{-4}}{5.7 \times 10^{-4}} \times 100\% = 41.58\%$
(c)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$
$Ka = 1.7 \times 10^{- 4}= \frac{x * x}{ 1.75 \times 10^{- 0}}$
$Ka = 1.7 \times 10^{- 4}= \frac{x^2}{ 1.75 \times 10^{- 0}}$
$ 2.97 \times 10^{- 4} = x^2$
$x = 1.72 \times 10^{- 2}$
Percent Ionization : $\frac{ 1.72 \times 10^{- 2}}{ 1.75} \times 100\% = 0.986\%$
