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 find the Ka of a monoprotic acid from its concentration and the pH of its solution, we follow these steps:
Given:
- Concentration of acid, [HA] = 0.0192 M
- pH = 2.53
- The acid is monoprotic, meaning it donates one proton per molecule.
Step 1: Calculate [H⁺] from pH
We use the formula:
$$
[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
$$
Step 2: Set up the equilibrium expression
The dissociation of a monoprotic acid is:
$$
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
$$
Initial concentrations:
- [HA] = 0.0192 M
- [H⁺] = 0
- [A⁻] = 0
Change in concentration:
- [HA] decreases by $x$
- [H⁺] and [A⁻] increase by $x$
At equilibrium:
- [HA] = 0.0192 – x
- [H⁺] = x
- [A⁻] = x
Since we calculated $[H^+] = 2.95 \times 10^{-3}$, we let:
$$
x = 2.95 \times 10^{-3}
$$
Step 3: Write the Ka expression
$$
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{x^2}{0.0192 – x}
$$
Plug in the values:
$$
K_a = \frac{(2.95 \times 10^{-3})^2}{0.0192 – 2.95 \times 10^{-3}} = \frac{8.70 \times 10^{-6}}{0.01625}
$$
$$
K_a \approx 5.35 \times 10^{-4}
$$
✅ Final Answer:
$$
\boxed{K_a \approx 5.35 \times 10^{-4}}
$$
The acid dissociation constant, Ka, is a quantitative measure of the strength of a weak acid in solution. It is derived from the equilibrium concentrations of the reactants and products in the dissociation of an acid in water. Since this problem involves a monoprotic acid, we assume it releases one proton (H⁺) per molecule of acid (HA) that dissociates.
We are given the molarity of the acid solution and the pH. The pH is a measure of hydrogen ion concentration and is calculated using the formula:
$$
\text{pH} = -\log[\text{H}^+]
$$
Reversing this formula allows us to determine the actual hydrogen ion concentration, which represents the amount of acid that has dissociated. With this, we assume that the concentration of A⁻ (the conjugate base) is equal to [H⁺] because of the 1:1 dissociation ratio of monoprotic acids.
Using this information, we apply the Ka expression, which is:
$$
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
$$
Here, the initial concentration of the acid is reduced by the amount that dissociates ($x$), and this amount is equal to the concentrations of H⁺ and A⁻ at equilibrium. By plugging in these values, we derive the numerical value for Ka.
The final result of $\boxed{5.35 \times 10^{-4}}$ tells us the extent to which the acid dissociates in water. This Ka value indicates a moderately weak acid, which partially dissociates, consistent with the given pH of 2.53. This method is commonly used in acid-base equilibrium calculations in chemistry.
