An aqueous solution of an unknown acid had a pH of 3.70.Titration of a 25.0 mL aliquot of the acid solution required 21.7 mL of 0.104 M aqueous sodium hydroxide for complete reaction.Assuming that the acid is monoprotic,what is its ionization constant? A)9.0 × 10^-2 B)2.0 × 10^-4 C)4.4 × 10^-7 D)3.6 × 10^-9 E)2.7 × 10^-11

Question

Quizplus · Accepted Answer

First, we can calculate the $[	ext{H}^+]$ concentration using the pH:

$$	ext{pH} = -\log[	ext{H}^+]$$
$$10^{-	ext{pH}} = [	ext{H}^+]$$
$$10^{-3.70} = [	ext{H}^+] = 2.0	imes10^{-4}	ext{ M}$$

Since we know that the acid is monoprotic, the reaction with sodium hydroxide must be:

$$	ext{HA + NaOH }ightarrow 	ext{ NaA + H}_2	ext{O}$$

We can write the balanced chemical equation and the expression for the equilibrium constant:

$$	ext{HA + NaOH }ightarrow 	ext{ NaA + H}_2	ext{O}$$
$$	ext{K}_	ext{a}=\frac{[	ext{A}^-][	ext{H}^+]}{[	ext{HA}]}$$

Before we start the calculations, we need to find the number of moles of $	ext{HA}$ in the aliquot. We can use the concentration and volume:

$$n = C 	imes V = 2.0	imes10^{-4}	ext{ M} 	imes 0.0250	ext{ L} = 5.00	imes10^{-6}	ext{ mol}$$

During the titration, we added 0.104 M sodium hydroxide, so the number of moles of $	ext{NaOH}$ added is:

$$n_{	ext{NaOH}} = 0.104	ext{ M} 	imes 0.0217	ext{ L} = 2.26	imes10^{-6}	ext{ mol}$$

Since the reaction is 1:1, the number of moles of $	ext{HA}$ that reacted is also $2.26	imes10^{-6}	ext{ mol}$. Therefore, the remaining number of moles of $	ext{HA}$ is:

$$n_{	ext{HA}} = 5.00	imes10^{-6}	ext{ mol} - 2.26	imes10^{-6}	ext{ mol} = 2.74	imes10^{-6}	ext{ mol}$$

At the equivalence point, all the moles of $	ext{HA}$ have reacted with $	ext{NaOH}$ and none remain:

$$[	ext{HA}] = 0	ext{ M}$$

The concentration of $	ext{NaOH}$ at the equivalence point can be calculated from the volume and concentration:

$$[	ext{NaOH}] = \frac{n_{	ext{NaOH}}}{V_{	ext{NaOH}}} = \frac{2.26	imes10^{-6}	ext{ mol}}{0.0217	ext{ L}} = 0.104	ext{ M}$$

Using this concentration and the $[	ext{H}^+]$ concentration we found earlier, we can calculate the concentration of $	ext{A}^-$. At equilibrium, we have the same number of moles of $	ext{A}^-$ as we had of $	ext{HA}$ before the titration:

$$[	ext{A}^-] = \frac{n_{	ext{HA}}}{V_{	ext{HA}}} = \frac{2.74	imes10^{-6}	ext{ mol}}{0.0250	ext{ L}} = 1.10	imes10^{-4}	ext{ M}$$

Finally, we can calculate the ionization constant:

$$	ext{K}_	ext{a}=\frac{[	ext{A}^-][	ext{H}^+]}{[	ext{HA}]} = \frac{(1.10	imes10^{-4}	ext{ M})(2.0	imes10^{-4}	ext{ M})}{2.74	imes10^{-6}	ext{ mol}} \approx 4.4	imes10^{-7}$$

Therefore, the answer is (C).

An Aqueous Solution of an Unknown Acid Had a PH