# The K_"a" for the weak acid HA is 4.0xx10^-6. what is the pH of a 0.01"M" solution of the acid? what is its "p"K_"a"?

Mar 30, 2016

$\text{p"K_"a} = 5.4$

$\text{pH} = 3.7$

#### Explanation:

$\text{p"K_"a}$ is simply $- {\log}_{10} \left({K}_{\text{a}}\right)$. It is a constant at any temperature and does not depend on the molarity of the solution.

"p"K_"a" = -log_10(K_"a")

$= - {\log}_{10} \left(4.0 \times {10}^{- 6}\right)$

$= 5.4$

To find the pH of the solution, we let the extent of dissociation be $x$.

Initially, before dissociation occurs, the concentration of each species, in $\text{M}$, are as follows.

• $\left[\text{HA}\right] = 0.01$
• $\left[{\text{H}}^{+}\right] = 0$
• $\left[{\text{A}}^{-}\right] = 0$

After $x$ amount of dissociation has occurred, the solution is in equilibrium. The concentration of each species, in $\text{M}$, are as follows.

• $\left[\text{HA}\right] = 0.01 - x$
• $\left[{\text{H}}^{+}\right] = x$
• $\left[{\text{A}}^{-}\right] = x$

Since they are in equilibrium, the relation below holds

K_"a" = frac{["H"^+] * ["A"^-]}{["HA"]}

$4.0 \times {10}^{- 6} = \frac{x \cdot x}{0.01 - x}$

This is a quadratic equation. But note that since ${K}_{\text{a}}$ is small, the acid is a weak acid. Therefore, dissociation occurs to a little extent only. Hence, we can assume $x \text{<<} 0.01$ to simplify our calculations.

$4.0 \times {10}^{- 6} = \frac{{x}^{2}}{0.01 - x}$

$\approx \frac{{x}^{2}}{0.01}$

Thus,

$x = \sqrt{\left(4.0 \times {10}^{- 6}\right) \times 0.01}$

$= 2.0 \times {10}^{- 4}$

Our original assumption of $x \text{<<} 0.01$ is valid and we can check that

$\frac{{\left(2.0 \times {10}^{- 4}\right)}^{2}}{0.01 - \left(2.0 \times {10}^{- 4}\right)} = 4.008 \times {10}^{- 6}$

$\approx 4.0 \times {10}^{- 6}$

Since we know $x$, we know that at equilibrium,

$\left[{\text{H}}^{+}\right] = x = 2.0 \times {10}^{- 4}$

The pH of the solution is given by

"pH" = -log_10(["H"^+])

$= - {\log}_{10} \left(2.0 \times {10}^{- 4}\right)$

$= 3.7$