Chapter 18

Q. 18.8

Determining Concentration and pH from K_a and Initial [HA]

Problem Propanoic acid (CH_3CH_2COOH, which we simplify to HPr) is a carboxylic acid whose salts are used to retard mold growth in foods. What are the [H_3O^+] and the pH of 0.10 M HPr (K_a  =  1.3×10^{−5})?


Verified Solution

Plan We know the initial concentration (0.10 M) and K_a  (1.3×10^{−5}) of HPr, and we need to find [H_3O^+] and pH. First, we write the balanced equation and the expression for K_a. We know [HPr]_{\text{init}} but not [HPr] (that is, the concentration at equilibrium). If we let x = [HPr]_{\text{dissoc}}, x is also [H_3O^+]_{\text{from HPr}} and [Pr^−] because each HPr dissociates into one H_3O^+ and one Pr^−. With this information, we set up a reaction table. We assume that, because HPr has a small K_a, it dissociates very little. After solving for x, which is [H_3O^+], we check the assumption. Then we use the value for [H_3O^+] to find the pH.

Solution Writing the balanced equation and expression for K_a:
            HPr(aq)  +  H_2O(l) \xrightleftharpoons[]{}  H_3O^+(aq)  +  Pr^−(aq)           K_a  =  \frac{[H_3O^+][Pr^−]}{[HPr]}  =  1.3×10^{−5}
Setting up a reaction table (Table 1), with x  =  [HPr]_{\text{dissoc}}  =  [H_3O^+]_{\text{from HPr}}  =  [Pr^−]  =  [H_3O^+]:

Making the assumption: K_a is small, so x is small compared with [HPr]_{\text{init}}; therefore, [HPr]_{\text{init}}  −  x  =  [HPr]  ≈  [HPr]_{\text{init}}, or 0.10 M − x ≈ 0.10 M. Substituting into the K_a expression and solving for x:

                 K_a  =  \frac{[H_3O^+][Pr^−]}{[HPr]}  =  1.3×10^{−5}  ≈  \frac{(x)(x)}{ 0.10}
                 x  ≈  \sqrt{(0.10)(1.3×10^{−5})}  =  1.1×10^{−3}  M  =  [H_3O^+]

Checking the assumption for [HPr]_{\text{dissoc}}:
            \frac{[H_3O^+]}{[HPr]_{\text{init}}}  ×  100  =  \frac{1.1×10^{−3}  M}{0.10  M}  ×  100  =  1.1% (<5%; assumption is justified.)
Finding the pH:

            pH  =  −\log  [H_3O^+]  =  −\log  (1.1×10^{−3})  =  2.96
Check The [H_3O^+] and pH seem reasonable for a dilute solution of a weak acid with a moderate K_a. By reversing the calculation, we can check the math: (1.1×10^{−3})^2/0.10  =  1.2×10^{−5}, which is within rounding of the given K_a.

Comment In Chapter 17 we introduced a benchmark, aside from the 5% rule, to see if the assumption is justified (see the discussion following Sample Problem 17.9):

· If \frac{[HA]_{\text{init}}}{K_a }  >  400, the assumption is justified: neglecting x introduces an error <5%.
· If \frac{[HA]_{\text{init}}}{K_a}  <  400, the assumption is not justified; neglecting x introduces an error >5%, so we solve a quadratic equation to find x.

In this sample problem, we have \frac{0.10}{1.3×10^{−5}}  =  7.7  ×  10^3, which is greater than 400. The alternative situation occurs in the next follow-up problem.

Table 1

Concentration (M) \mathbf{HPr(aq)    +     H_2O(l)        \xrightleftharpoons[]{}      H_3O^+(aq)      +        Pr^−(aq)}
Initial 0.10                              —                                  0                                    0
Change −x                                 —                                 +x                                 +x
Equilibrium 0.10 − x                       —                                  x                                     x