para-Aminobenzoic acid, PABA, is a powerful sunscreening agent whose salts are used widely in sun tanning and screening lotions. The parent acid, which we may symbolize as H-Paba, is a weak acid with a pka of 4.92. What will be the [H+] and pH of a 0.030 M solution of this acid? I need help solving this.

Answers ( if you want to check ):

pH = 3.22
[H+] = 6.0 x 10^-4 mol/l

Start with the definition of Ka and pKa.

pKa = -log(K_a).

Ka = 10^{-pKa} = 10^{-4.92} = 1.202 \times 10^{-5}

For a general acid, Ka is defined

Ka = \frac {[A^-][H^+]}{[HA]} where [] indicates the concentration.

Ka =\frac {[PABA^-][H^+]}{[HPABA]} = 1.20 \times 10^{-5} \frac {mol}{L} (Equation 1)

(Note the units of Ka and the units of the concentrations.)

At first it looks like we have one equation and three unknowns. However, remember that this comes from the following equillibrium

HPABA \rightleftharpoons H^+ + PABA^-

if we start with pure HPABA, whose concentration is C (0.030 M), then

[PABA] = C - x where x is the amount that's dissociated and (Equation 2)

[H^+] = [PABA^-] = x (Equation 3)

So we have three equations and three unknowns. Substituting into equation 1, we have

\frac {x^2}{0.03-x} = 1.2 \times 10^{-5}

rearranging

x^2 + 1.20 \times 10^{-5}x - 3.6 \times 10^{-7} = 0

Solving yields x=6.060 \times 10^{-4} (we ignore the negative root)

Therefore [H^+] =6.060 \times 10^{-4} \,and\,pH=3.218

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For very weak acids, many people ignore the x in (C-x), and just calculate from there:

If x is small compared to C,

\frac {x^2}{C} = \frac {x^2}{0.03} = 1.2 \times 10^{-5}

x = \sqrt {3.6\times 10^{-7}} = 6.000 \times 10^{-4} and we have almost the same answer (pH =3.222) .

Thanks a lot for the help... I just want to know when to use the quadratic formula or just drop the x's because sometimes I wouldn't know if it's a weak acid. Are there any other ways to find out?

If you drop the x in the denominator, you can get an estimate of x. If it's small compared to C, you're finished.