How do you calculate pH from acid dissociation constant?

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How do you calculate pH from acid dissociation constant?

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Answer 1 · 2014-03-19T00:18:23

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 $[H^{+}]$ $[H^+]$ .

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

$H O A C \leftrightarrow H^{+} + O A c^{-}$ $HOAC ↔ H^+ + OAc^-$
$K_{a} = 1.8 x 10^{- 5} = \frac{[H^{+}] [O A c^{-}]}{[H O A C]}$ $K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC])$
If the acid is weak, then only a small concentration, $x$ $x$ , will dissociate. We can rewrite the equation as
$K_{a} = 1.8 x 10^{- 5} = \frac{[H^{+}] [O A c^{-}]}{[H O A C]}$ $K_a = 1.8x10^(-5) = ([H^+][OAc^-])/([HOAC])$ = $\frac{x^{2}}{0.2 - x}$ $(x^2)/(0.2-x)$ & approximately; $\frac{x^{2}}{0.2}$ $x^2/0.2$
where we have ignored $x$ $x$ 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...
$x = {(0.2 \cdot 1.8 x 10^{- 5})}^{\frac{1}{2}}$ $x = (0.2 · 1.8x10^(-5))^(1/2)$ = $1.9 x 10^{- 3}$ $1.9x10^(-3)$ = $[H^{+}]$ $[H^+]$
pH = $- log ([H^{+}])$ $-log([H^+])$ = 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.

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How do you calculate pH from acid dissociation constant?

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