Question

How do you calculate pH from acid dissociation constant?

Answer 1

For a weak acid, set up the equilibrium expression for dissociation to ions in solution, then solve this equation for the hydronium ion concentration. The pH can be calculated directly from `[H^+]`.

Example: The pH of 0.2 M acetic acid (HOAc)

`HOAC ↔ H^+ + OAc^-`
`K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC])`
If the acid is weak, then only a small concentration, `x`, will dissociate. We can rewrite the equation as
`K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC])` = `(x^2)/(0.2-x)` & approximately; `x^2/0.2`
where we have ignored `x` in the denominator because it is a small number compared with 0.2. Solving this equation gives * See note below about what to do if you do not want to ignore this...
`x = (0.2 · 1.8x10^(-5))^(1/2)` = `1.9x10^(-3)` = `[H^+]`
pH = `-log([H^+])` = 2.72

NOTE: If you don't simpify 0.2-x = 0.2, you need to use the quadratic equation to solve for the pH. Here is a video which discusses how to do this.