Answer
(a) 3.31%
(b) 31.3%
(c) 75.4%
- When the initial acid concentration decreases, the percent ionization increases.
Work Step by Step
$Ka (HF) = 6.6 \times 10^{-4}$
- Drawing the ICE table, we get:
-$[H_3O^+] = [A^-] = x$
-$[HA] = [HA]_{initial} - x$
For approximation, we consider: $[HA] = [HA]_{initial}$
(a)
Use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 6 \times 10^{- 1}}$
$Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 6 \times 10^{- 1}}$
$ 3.95 \times 10^{- 4} = x^2$
$x = 1.98 \times 10^{- 2}$
%Ionization: $\frac{ 1.98 \times 10^{- 2}}{ 6 \times 10^{- 1}} \times 100\% = 3.31\%$
--------
(b)
Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 4.6 \times 10^{- 3}}$
$Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 4.6 \times 10^{- 3}}$
$ 3.03 \times 10^{- 6} = x^2$
$x = 1.74 \times 10^{- 3}$
Ionization: $\frac{ 1.74 \times 10^{- 3}}{ 4.59 \times 10^{- 3}} \times 100\% = 37.8\%$
Ionization > 5%, so we to consider the "x" of the acid concentration:
$Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 4.6 \times 10^{- 3}- x}$
$ 3.03 \times 10^{- 6} - 6.6 \times 10^{- 4}x = x^2$
$ 3.03 \times 10^{- 6} - 6.6 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 6.6 \times 10^{- 4})^2 - 4 * (-1) *( 3.03 \times 10^{- 6})$
$\Delta = 4.35 \times 10^{- 7} + 1.21 \times 10^{- 5} = 1.25 \times 10^{- 5}$
$x_1 = \frac{ - (- 6.6 \times 10^{- 4})+ \sqrt { 1.25 \times 10^{- 5}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 6.6 \times 10^{- 4})- \sqrt { 1.25 \times 10^{- 5}}}{2*(-1)}$
$x_1 = - 2.1 \times 10^{- 3} (Negative)$
$x_2 = 1.44 \times 10^{- 3}$
Ionization: $\frac{ 1.44 \times 10^{- 3}}{ 4.6 \times 10^{-3}} \times 100\% = 31.3\%$
-----
(c)
Use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][A^-]}{ [HA]}$
$Ka = 6.6 \times 10^{- 4}= \frac{x * x}{ 2.8 \times 10^{- 4}}$
$Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 2.8 \times 10^{- 4}}$
$ 1.84 \times 10^{- 7} = x^2$
$x = 4.29 \times 10^{- 4}$
5% test: $\frac{ 4.29 \times 10^{- 4}}{ 2.8 \times 10^{- 4}} \times 100\% = 153\%$
Ionization > 5%, so we to consider the "x" of the acid concentration:
$Ka = 6.6 \times 10^{- 4}= \frac{x^2}{ 2.8 \times 10^{- 4}- x}$
$ 1.84 \times 10^{- 7} - 6.6 \times 10^{- 4}x = x^2$
$ 1.84 \times 10^{- 7} - 6.6 \times 10^{- 4}x - x^2 = 0$
Bhaskara:
$\Delta = (- 6.6 \times 10^{- 4})^2 - 4 * (-1) *( 1.84 \times 10^{- 7})$
$\Delta = 4.35 \times 10^{- 7} + 7.39 \times 10^{- 7} = 1.17 \times 10^{- 6}$
$x_1 = \frac{ - (- 6.6 \times 10^{- 4})+ \sqrt { 1.17 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 6.6 \times 10^{- 4})- \sqrt { 1.17 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 8.71 \times 10^{- 4} (Negative)$
$x_2 = 2.11 \times 10^{- 4}$
Ionization: $\frac{ 2.11 \times 10^{- 4}}{ 2.8 \times 10^{- 4}} \times 100\% = 75.4\%$