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 acid dissociation constant (Ka) for a monoprotic weak acid, we use the pH to determine the concentration of H⁺ ions, then apply the expression for Ka using the ICE table method.
Step 1: Use pH to find [H⁺]
The pH of the solution is given as 2.53. We use the formula: [H+]=10−pH=10−2.53[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} [H+]≈2.95×10−3 M[\text{H}^+] \approx 2.95 \times 10^{-3} \text{ M}
Step 2: Write the dissociation equation
Since it’s a monoprotic acid (denoted as HA), the reaction is: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
Initial concentration of HA = 0.0192 M
Change in [H⁺] = +2.95 × 10⁻³ M
Change in [A⁻] = +2.95 × 10⁻³ M
Change in [HA] = –2.95 × 10⁻³ M
Step 3: Set up the Ka expression
Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
Substitute known values:
- [H+]=2.95×10−3[H^+] = 2.95 \times 10^{-3}
- [A−]=2.95×10−3[A^-] = 2.95 \times 10^{-3}
- [HA]=0.0192−2.95×10−3=0.01625[HA] = 0.0192 – 2.95 \times 10^{-3} = 0.01625
Ka=(2.95×10−3)20.01625K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} Ka=8.70×10−60.01625K_a = \frac{8.70 \times 10^{-6}}{0.01625} Ka≈5.35×10−4K_a \approx 5.35 \times 10^{-4}
✅ Final Answer:
Ka≈5.35×10−4\boxed{K_a \approx 5.35 \times 10^{-4}}
Explanation (300+ words):
In acid-base chemistry, the Ka (acid dissociation constant) quantifies the strength of a weak acid. For monoprotic acids (which donate one proton), the dissociation in water is represented by: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
We’re given the initial concentration of the acid solution as 0.0192 M and a pH of 2.53. Since pH is the negative logarithm of the hydrogen ion concentration, we first determine [H⁺] by reversing the logarithmic function: [H+]=10−pH=10−2.53≈2.95×10−3 M[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \text{ M}
This value also represents the concentration of A⁻, because for every mole of HA that dissociates, one mole each of H⁺ and A⁻ is produced.
We then use an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations. Initially, HA is at 0.0192 M, and no H⁺ or A⁻ is present. As dissociation occurs, HA decreases by x (which is 2.95 × 10⁻³), and H⁺ and A⁻ each increase by x.
With equilibrium concentrations known, we plug values into the Ka expression: Ka=[H+][A−][HA]=(2.95×10−3)20.01625=5.35×10−4K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(2.95 \times 10^{-3})^2}{0.01625} = 5.35 \times 10^{-4}
This value reflects a moderate acid strength: not too weak, but not strong either. It shows a small but significant degree of ionization, which is typical for weak acids.