# The Ka value of a weak acid, HA, is 10-6. An initial solution containing 1.0 M of HA is prepared. What is the pH of the final solution at equilibrium?

Aug 7, 2017

$\textsf{p H = 3}$

#### Explanation:

For a weak acid you can use the expression:

sf(pH=1/2[pK_a-log[acid])

$\therefore$sf(pH=1/2[6-log(1)]

$\textsf{p H = 3}$

Aug 7, 2017

$\text{pH} = 3$.

If you wish, you could use the equation shown at the end here, which is what Michael has done.

For this weak acid, a general equilibrium can be written:

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

After you make your ICE table, or whatever floats your boat, you should end up with a mass action expression of:

${K}_{a} = {10}^{- 6} = \frac{{x}^{2}}{1.0 - x}$

For typical concentrations (about $0.01 - \text{1.0 M}$), if ${K}_{a} < {10}^{- 5}$ or so, we can consider the small $x$ approximation without much loss of accuracy.

(Furthermore, the percent dissociation decreases with increasing starting concentration, so a $\text{1.0 M}$ concentration is even better for the approximation than, say, $\text{0.01 M}$.)

This says that we can write:

${K}_{a} \approx \frac{{x}^{2}}{1.0}$

or that

x -= ["H"^(+)] = sqrt(K_a("1.0 M"))

$= \sqrt{{10}^{- 6} \cdot 1}$

$= {10}^{- 3}$ $\text{M}$

And thus, the $\text{pH}$ would be:

$\textcolor{b l u e}{{\text{pH") = -log["H}}^{+}}$

$= - \log \left({10}^{- 3}\right) = \textcolor{b l u e}{3}$