Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.
The Correct Answer and Explanation is:
To calculate the Ka (acid dissociation constant) for a monoprotic acid, we use the relationship between pH, concentration, and the expression for Ka:
Given:
- Initial concentration of the acid: $[HA]_0 = 0.0192 \, \text{M}$
- pH of the solution: $2.53$
- The acid is monoprotic (donates one proton per molecule).
Step 1: Calculate $[H^+]$ from pH
Use the formula:
$$
[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
$$
Step 2: Set up ICE Table
$$
\begin{array}{c|ccc}
& \text{HA} & \rightleftharpoons & \text{H}^+ & + & \text{A}^- \
\hline
\text{Initial (M)} & 0.0192 & & 0 & & 0 \
\text{Change (M)} & -x & & +x & & +x \
\text{Equilibrium (M)} & 0.0192 – x & & x & & x \
\end{array}
$$
From step 1: $x = [H^+] = 2.95 \times 10^{-3} \, \text{M}$
Step 3: Use Ka Expression
$$
Ka = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{0.0192 – x}
$$
Substitute values:
$$
Ka = \frac{(2.95 \times 10^{-3})^2}{0.0192 – 2.95 \times 10^{-3}} = \frac{8.70 \times 10^{-6}}{0.01625}
$$
$$
Ka \approx 5.35 \times 10^{-4}
$$
✅ Final Answer:
$$
\boxed{5.35 \times 10^{-4}}
$$
Explanation (300+ words):
The problem involves calculating the acid dissociation constant (Ka) for a monoprotic acid, which dissociates in water as follows:
$$
HA \rightleftharpoons H^+ + A^-
$$
The Ka quantifies how much of the acid dissociates. A stronger acid has a higher Ka because more $H^+$ ions are produced.
We are given the initial concentration of the acid and the pH. First, we convert the pH to $[H^+]$ using the formula:
$$
[H^+] = 10^{-\text{pH}} = 10^{-2.53}
$$
This gives the amount of acid that has dissociated into $H^+$ (and $A^-$, in equal amounts for a monoprotic acid). We assume $x = [H^+] = 2.95 \times 10^{-3} \, \text{M}$, meaning 0.00295 M of the acid has dissociated.
Now, we calculate how much of the acid remains: $0.0192 – x = 0.01625 \, \text{M}$. The Ka formula for this dissociation is:
$$
Ka = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{[HA]_{\text{remaining}}}
$$
By plugging in the known values:
$$
Ka = \frac{(2.95 \times 10^{-3})^2}{0.01625} = 5.35 \times 10^{-4}
$$
This result shows the acid is a weak acid because its Ka is less than 1. If all the acid had dissociated, it would be a strong acid, and the pH would be much lower. This calculation method—using an ICE table and pH to find Ka—is standard in acid-base equilibrium problems and is especially important when the acid is weak and only partially dissociates.