## 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