# The K_a of a monoprotic weak acid is 4.67 x 10^-3. What is the percent ionization of a 0.171 M solution of this acid?

May 9, 2017

$\text{% dissociation = 15%}$

#### Explanation:

We address the equilibrium,

$H A \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + {A}^{-}$

And ${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]} = 4.67 \times {10}^{-} 3$

Now if $x \cdot m o l \cdot {L}^{-} 1$ $H A$ dissociates then..........

$4.67 \times {10}^{-} 3 = {x}^{2} / \left(0.171 - x\right)$

And this is quadratic in $x$, but instead of using the tedious quadratic equation, we assume that $0.171 \text{>>} x$, and so.......

$4.67 \times {10}^{-} 3 \cong {x}^{2} / \left(0.171\right)$

And thus ${x}_{1} = \sqrt{4.67 \times {10}^{-} 3 \times 0.171} = 0.0283 \cdot m o l \cdot {L}^{-} 1$, and now we have an approx. for $x$, we can recycle this back into the expression:

${x}_{2} = 0.0258 \cdot m o l \cdot {L}^{-} 1$

${x}_{3} = 0.0260 \cdot m o l \cdot {L}^{-} 1$

${x}_{4} = 0.0260 \cdot m o l \cdot {L}^{-} 1$

Since the values have converged, I am willing to accept this answer. (This same answer would have resulted from the quadratic equation, had we bothered to do it.)

And thus percentage dissociation................

="Concentration of hydronium ion"/"Initial concentration of acid"xx100%=(0.0260*mol*L^-1)/(0.171*mol*L^-1)xx100%=15%