# Question 742ba

Feb 14, 2018

Unknown without data...

#### Explanation:

We interrogate the equilibrium....

HOAc(aq)+H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc

$H O A c = {H}_{3} C - C \left(= O\right) O H$

Now it is a fact that ${K}_{a} \left(H O A c\right) = {10}^{- 4.76} = 1.74 \times {10}^{-} 5$..

And so we set up the equilibrium expression...

1.74xx10^(-5)=([H_3O^+][""^(-)OAc])/([HOAc])

And if we put $\left[{H}_{3} {O}^{+}\right] = x$...then by stoichiometry...

$1.74 \times {10}^{- 5} = \frac{{x}^{2}}{0.0015 - x}$..

And if $x \text{<<} 0.0015$...then...$x \cong \sqrt{1.74 \times {10}^{-} 5 \times 0.0015}$

${x}_{1} = 1.60 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...

And we can use this value for a second approximation....

${x}_{2} = 1.53 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...

${x}_{3} = 1.53 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...

...and since the approximations have converged, I am prepared to accept this as the true value...

And so "% ionization"="moles of anion"/"moles of starting acid"xx100%

=(1.53xx10^-4*mol*L^-1)/(0.0015*mol*L^-1)xx100%=10.2%#

This $\text{% ionization}$ is GREATER than for a more concentrated solution of weak acid .... at increasing dilution $\text{% ionization}$ becomes greater for weak acids....