# If enough of a monoprotic acid is dissolved in water to produce a 0.0150 M solution with a pH of 6.05, what is the equilibrium constant tor the acid?

Nov 22, 2015

${K}_{a} = 5.3 \cdot {10}^{- 11}$

#### Explanation:

The idea here is that you need to use the pH of the solution to determine the concentration of hydronium ions, ${\text{H"_3"O}}^{+}$, that is required in order for the solution to have a pH of $6.05$.

Even without doing any calculations, you can say that the acid dissociation constant, ${K}_{a}$, for this monoprotic weak acid will be very, very small.

The pH of the resulting solution is relatively close to the pH of pure water at room temperature, which means that a very, very small number of acid molecules will actually ionize.

To find the value of the acid dissociation constant, use an ICE table using a generic $\text{HA}$ monoprotic acid

${\text{ ""HA"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A}}_{\textrm{\left(a q\right]}}^{-}$

color(purple)("I")" " " " "0.0150" " " " " " " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " " " " " "(+x)" " " " " "(+x)
color(purple)("E")" "(0.0150-x)" " " " " " " " " " " "x" " " " " " " " " "x

By definition, ${K}_{a}$ will be equal to

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

Now use the pH of the solution to determine the equilibrium concentration of hydronium ions

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

["H"_3"O"^(+)] = 10^(-6.05) = 8.91 * 10^(-7)"M"

As you can see from the ICE table, the equilibrium concentration of hydronium ions, which is equal to that of the acid's conjugate base ${\text{A}}^{-}$, is equal to $x$.

This means that the expression for ${K}_{a}$ becomes

${K}_{a} = \frac{x \cdot x}{0.0150 - x} = {\left(8.91 \cdot {10}^{- 7}\right)}^{2} / \left(0.0150 - 8.91 \cdot {10}^{- 7}\right)$

The denominator can be approximated with

$0.0150 - 8.91 \cdot {10}^{- 7} \approx 0.0150$

This means that ${K}_{a}$ will be equal to

${K}_{a} = \frac{{8.91}^{2} \cdot {10}^{- 14}}{0.0150} = \textcolor{g r e e n}{5.3 \cdot {10}^{- 11}}$

Indeed, like we initially predicted, you're dealing with a very weak acid.