Question

if the ka of a monoprotic weak acid is 3.2x10^-6 what is the pH of a 0.15 M solution of this acid

Answer 1

You need an ICE chart, since this is a weak acid:
\[\color{green}{HA \rightleftharpoons H^+ + A^-\\0.15 \quad ~~~~~0 \qquad ~0\\ -x\quad ~+x \quad +x\\ 0.15 - x ~~x \quad ~~~~x}\]
Since Ka is small, 0.15 - x \( \approx\) 0.15 (hundred rule)

Now,
\[\begin{align} \color{green}{K_a = \frac{[H^+][A^-]}{[HA]} \\ 3.2 \times 10^{-6} = \frac{x^2}{(0.15 - x)} \\ 3.2 \times 10^{-6} \approx \frac{x^2}{0.15}} \end{align}\]

Solve for x. Then, once you have found the equilibrium concentrations, find the reaction quotient, Q. If it is approximately equal to Ka you know that the simplifying assumption was valid.

Answer 2

thank you!