# A 0.01 M solution of an acid has a pH = 5.0. What is the Ka value?

May 30, 2017

${K}_{a} = 1 \cdot {10}^{- 8}$

#### Explanation:

I'm assuming that you're working with a monoprotic weak acid here so that the ionization equilibrium can be written like this

${\text{HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "A"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

Now, you know that the solution has

$\text{pH} = 5$

As you know, the $\text{pH}$ of the solution is defined as

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

You can rearrange this equation to find the equilibrium concentration of hydronium cations.

log(["H"_3"O"^(+)]) = - "pH"

This is equivalent to

10^log(["H"_3"O"^(+)]) = 10^(-"pH")

which gets you

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

In your case, you will have

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 5.0} = 1.0 \cdot {10}^{- 5}$ $\text{M}$

Now, notice that every mole of $\text{HA}$ that ionizes produces $1$ mole of ${\text{A}}^{-}$, the conjugate base of the acid, and $1$ mole of hydronium cations.

This means that, at equilibrium, the solution has

["A"^(-)] = ["H"_3"O"^(+)] " "-> produced in a $1 : 1$ mole ratio

$\left[{\text{A}}^{-}\right] = 1.0 \cdot {10}^{- 5}$ $\text{M}$

The initial concentration of the acid will decrease because some of the molecules ionize to produce ${\text{A}}^{-}$ and ${\text{H"_3"O}}^{+}$.

So, in order for the ionization to produce $\left[{\text{H"_3"O}}^{+}\right]$, the initial concentration of the acid must decrease by $\left[{\text{H"_3"O}}^{+}\right]$.

This means that, at equilibrium, the concentration of the weak acid will be equal to

$\left[{\text{HA"] = ["HA"]_"initial" - ["H"_3"O}}^{+}\right]$

In your case, you will have

["HA"] = "0.01 M" - 1.0 * 10^(-5)color(white)(.)"M"

["HA"] = "0.00999 M"

By definition, the acid dissociation constant, ${K}_{a}$, will be equal to

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

Plug in your values to find

${K}_{a} = \left(1.0 \cdot {10}^{- 5} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))) * 1.0 * 10^(-5)color(white)(.)"M")/(0.00999color(red)(cancel(color(black)("M}}}}\right)$

${K}_{a} = 1.001 \cdot {10}^{- 8}$ $\text{M}$

Rounded to one significant figure and expresses without added units, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{a} = 1 \cdot {10}^{- 8}}}}$