We have a solution $\ce{CH3COOH}$ (acetic acid) with $c=0.02~\mathrm{mol/L}$ and $K_\mathrm a(\ce{CH3COOH}) = 1.8\cdot10^{-5}$. Calculate the $\mathrm{pH}$ of this solution.

All I know is that it is a weak acid and that $$\mathrm{pH}= \mathrm{p}K_\mathrm a + \log\left(\frac{B}{A}\right).$$ I know $A$ but there is no information about $B$.


This type of problem is best solved using the ICE (Initial Change Equilibrium) method.

First, write down the chemical equation and the law of mass action for this reaction.

$$\ce{CH3COOH <=> H+ + CH3COO-}$$ $$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$$

Second, determine the initial concentration of each of the three species. Make a little table to keep everything straight:

\begin{array}{lc} \hline &\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline \text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \hline \end{array}

At equilibrium, some of the acetic acid will be ionized. How much? We don't know right now, so let's have that be a variable; $x$. At equilibrium, the concentration of acetic acid will have decreased by $x$. How much will the concentration of $\ce{H+}$ and acetate anion increase? Since the stoichiometric coefficient of all species is 1, then those concentrations will increase by $x$. Let's add that to our table. Let's also write an expression for the equilibrium concentrations in terms of the initial concentrations and $x$.

\begin{array}{lc} \hline &\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline \text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \text{C}& -x &&+x &&+x \\ \text{E}& 0.02-x &&x &&x \\ \hline \end{array}

Now, plug these expressions into the law of mass action:

$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{(x)(x)}{(0.02-x)}$$

As long as $x\ne 0.02$, which is very unlikely at equilibrium, we have a solvable quadratic equation:

$$K_a=1.8\times 10^{-5}= \dfrac{(x)(x)}{(0.02-x)}$$ $$3.6\times 10^{-7}-1.8\times 10^{-5}x-x^2=0$$

Some texts will encourage you to make the simplification $0.02-x \approx 0.02$ since $x<<0.02$, however, WolframAlpha can solve the original equation (and so can you).

The equation has two roots: $x=5.91\times 10^{-4}$ and $x=-6.09\times 10^{-4}$. The second one is nonsense (we cannot have negative concentrations).

If you make the assumption that $0.02−x \approx 0.02$, then the positive root is $x=6.0 \times 10^{−4}$, not a big difference given the significant figures. If you knew your initial concentration more precisely, this assumption would be invalid.

Now, we know what $x$ is. What do we still need to know to determine pH? How is $x$ related to pH?


@bennorris method works, but I think my method may save some time.

\begin{array}{lc} \hline &\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline \text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \text{C}& -x &&+x &&+x \\ \text{E}& 0.02-x &&x &&x \\ \hline \end{array}

I keep the acetic acid concentration to be the same because I assume the x value will be really small( you can prove it later)

$$K_a = 1.8\times 10^{-5}=\dfrac{[\ce{H+}][\ce{C2H3O2-}]}{[\ce{HC2H3O2}]} = \dfrac{(x)(x)}{(0.02)}$$

$$ x=6.0 \times 10^{-4}$$

You can prove the assumption by dividing your x value with your initial concentration of acetic acid $$ \dfrac{[\pu{6\times 10^{-4} M}]}{[\pu{0.02 M}]} =0.03 =3\% < 5\%$$ If the percentage you get is smaller than five percent, your assumption will be correct. This method will save you some time as you don't have to deal withe the quadratic equation and give you a approximate answer

  • 2
    $\begingroup$ I have to say that if you turned in this answer in an exam I was marking you would score zero. $\endgroup$ – John Rennie Nov 9 '17 at 9:12
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    $\begingroup$ Yes, but the ICE table should initially be setup as Ben did it. Then make the assumption that $0.02-x \approx 0.02$. The 5% criteria seems to have been pulled from thin air. I'd base the needed precision on significant figures (2) giving a need for about 1%. Lastly the acetate species could have been normalized to 100% which would have given a result very close to the "true" answer. $\endgroup$ – MaxW Nov 12 '17 at 21:40

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