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 acid, we need to use the information provided about the concentration of the acid and the pH of the solution. Here’s a step-by-step approach:
Step 1: Write the dissociation reaction for the monoprotic acid.
The monoprotic acid (HA) dissociates in water as follows:
[
HA \rightleftharpoons H^+ + A^-
]
Step 2: Use the pH to calculate the concentration of ( H^+ ) ions.
We are given that the pH of the solution is 2.53. The pH is related to the concentration of hydrogen ions (( [H^+] )) by the following equation:
[
\text{pH} = -\log[H^+]
]
Solving for ( [H^+] ):
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
]
Step 3: Set up an ICE table to track the dissociation of the acid.
Let’s assume that the initial concentration of HA is ( [HA] = 0.0192 \, \text{M} ). For simplicity, we will define the change in concentration of HA that dissociates as ( x ). Therefore, at equilibrium, the concentrations of the species will be:
| Species | Initial Concentration | Change in Concentration | Equilibrium Concentration |
|---|---|---|---|
| HA | 0.0192 M | -x | 0.0192 – x |
| ( H^+ ) | 0 | +x | x |
| A⁻ | 0 | +x | x |
We know from the pH calculation that at equilibrium, ( [H^+] = 2.95 \times 10^{-3} \, \text{M} ). Therefore, ( x = 2.95 \times 10^{-3} ).
Step 4: Use the expression for the acid dissociation constant, Ka.
The Ka expression for the dissociation of HA is:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substituting the values at equilibrium:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}}
]
Calculating the denominator:
[
0.0192 – 2.95 \times 10^{-3} = 0.01625
]
Now, calculate the Ka:
[
K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} = \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.36 \times 10^{-4}
]
Final Answer:
The acid dissociation constant ( K_a ) for the monoprotic acid is approximately:
[
K_a \approx 5.36 \times 10^{-4}
]
Explanation:
To find the Ka, we used the pH of the solution to determine the concentration of hydrogen ions. By assuming that the acid dissociates according to the monoprotic acid dissociation reaction, we applied the ICE (Initial, Change, Equilibrium) table to track the concentration changes during dissociation. We then used the Ka expression to calculate the acid dissociation constant based on the concentrations at equilibrium.