# How do you calculate pH from acid dissociation constant?

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 can be calculated directly from ##[H^+]##.

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

##HOAC &harr; H^+ + OAc^-##
##K_a = 1.8×10^(-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 = 1.8×10^(-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 &middot; 1.8×10^(-5))^(1/2)## = ##1.9×10^(-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.

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