# If the K_a of a monoprotic weak acid is 8.1 times 10^-6, what is the pH of a 0.33 M solution of this acid?

May 10, 2017

$p H = 2.79$

#### Explanation:

$\left(i\right)$ We first write a stoichiometric equation to inform our reasoning:

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

$\left(i i\right)$ And then we write the equilibrium expression:

$\frac{\left[{H}_{3} {O}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]} = {K}_{a} = 8.1 \times {10}^{-} 6$

$\left(i i i\right)$ And then we solve the problem using approximations IF APPROPRIATE.

If the amount of dissociation is $x \cdot m o l \cdot {L}^{-} 1$, then our expression becomes:

${x}^{2} / \left(0.33 - x\right) = 8.1 \times {10}^{-} 6$

If $0.33 \text{>>>} x$, then $0.33 - x \cong 0.33$; and we must justify this approx. later. Solving for $x = \sqrt{8.1 \times {10}^{-} 6 \times 0.33}$, we gets....

${x}_{1} = 1.63 \times {10}^{-} 3$; a value that is indeed small compared to $0.33$. Just to belabour the point, we can input our first approx. back in the calculation, and get a 2nd approx.......

${x}_{2} = 1.63 \times {10}^{-} 3$

Since the approximations have converged, we are willing to accept this value as the true value (i.e. the same as if we exactly solved the quadratic in $x$).

Given $x = 1.63 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1 = \left[{H}_{3} {O}^{+}\right]$, $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} 1.63 \times {10}^{-} 3 = 2.79$.

What is $p O H$ of this solution?