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 understand the relationship between the concentrations of the acid, its dissociation products, and the pH of the solution.
Step-by-Step Solution
- Write the dissociation equation for the monoprotic acid (HA):
[
HA \rightleftharpoons H^+ + A^-
] - Given Information:
- The concentration of the acid (( [HA] )) is 0.0192 M.
- The pH of the solution is 2.53.
- Calculate the concentration of ( H^+ ) ions using the pH:
The pH is related to the concentration of ( H^+ ) ions by the equation:
[
\text{pH} = -\log[H^+]
]
Given the pH is 2.53, we can find the concentration of ( H^+ ) as follows:
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
] - Set up an ICE table to track the dissociation of the acid: Species Initial Concentration (M) Change in Concentration (M) Equilibrium Concentration (M) ( HA ) 0.0192 -( x ) 0.0192 – ( x ) ( H^+ ) 0 +( x ) ( x ) ( A^- ) 0 +( x ) ( x ) From the ICE table, we know that at equilibrium, the concentration of ( H^+ ) is ( x ), which we calculated as ( 2.95 \times 10^{-3} \, \text{M} ). Therefore, at equilibrium:
[
[H^+] = 2.95 \times 10^{-3} \, \text{M}
]
and
[
[A^-] = 2.95 \times 10^{-3} \, \text{M}
] The concentration of the acid at equilibrium is:
[
[HA] = 0.0192 – 2.95 \times 10^{-3} \approx 0.0163 \, \text{M}
] - Write the expression for the acid dissociation constant (Ka):
The acid dissociation constant ( K_a ) is given by:
[
K_a = \frac{[H^+][A^-]}{[HA]}
] - Substitute the equilibrium concentrations into the Ka expression:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0163}
]
[
K_a = \frac{8.70 \times 10^{-6}}{0.0163}
]
[
K_a \approx 5.34 \times 10^{-4}
]
Final Answer:
The Ka for the acid is approximately 5.34 × 10⁻⁴.
Explanation:
This calculation involves applying the concepts of acid dissociation and equilibrium. The dissociation of the monoprotic acid into ( H^+ ) and ( A^- ) establishes an equilibrium. By using the pH to determine the concentration of ( H^+ ), we can track how much of the acid dissociates. The Ka expression is then used to relate the equilibrium concentrations of the acid and its dissociation products to the acid dissociation constant, giving us the strength of the acid. The result indicates a moderately weak acid, as the Ka is less than 1 but still significant.