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 the monoprotic acid, we need to use the information provided, including the molarity of the acid solution (0.0192 M) and the pH of the solution (2.53). Here’s a step-by-step process for calculating Ka:
Step 1: Calculate the concentration of hydrogen ions ([H+])
The pH of a solution is related to the concentration of hydrogen ions by the following equation: pH=−log[H+]\text{pH} = -\log[\text{H}^+]
Given the pH is 2.53, we can rearrange the equation to solve for [H+]: [H+]=10−pH=10−2.53=2.95×10−3 M[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} = 2.95 \times 10^{-3} \, \text{M}
Step 2: Set up the ICE table
The monoprotic acid dissociates as follows: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
Using an ICE (Initial, Change, Equilibrium) table:
| Species | Initial Concentration (M) | Change in Concentration (M) | Equilibrium Concentration (M) |
|---|---|---|---|
| HA | 0.0192 | -x | 0.0192 – x |
| H+ | 0 | +x | 2.95 × 10⁻³ |
| A- | 0 | +x | 2.95 × 10⁻³ |
Here, xx represents the amount of acid dissociated, which is equal to the concentration of H+\text{H}^+ at equilibrium.
Thus, we can conclude that x=2.95×10−3 Mx = 2.95 \times 10^{-3} \, \text{M}.
Step 3: Solve for the Ka
The Ka expression for the dissociation of the acid is: Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
At equilibrium: Ka=(2.95×10−3)(2.95×10−3)0.0192−2.95×10−3=8.70×10−60.01625=5.36×10−4K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}} = \frac{8.70 \times 10^{-6}}{0.01625} = 5.36 \times 10^{-4}
Thus, the Ka for the monoprotic acid is: Ka=5.36×10−4K_a = 5.36 \times 10^{-4}
Conclusion
The acid dissociation constant (Ka) for the monoprotic acid is 5.36×10−45.36 \times 10^{-4}.
This calculation involves understanding the relationship between pH and the concentration of hydrogen ions and applying the equilibrium expression for acid dissociation.