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 acid dissociation constant (( K_a )) for the monoprotic acid, we follow these steps:
Step 1: Define the given values
- Initial concentration of the acid: ( [HA] = 0.0192 ) M
- pH of the solution: ( 2.53 )
Step 2: Determine ( [H^+] )
Since pH is given, we can find the hydronium ion concentration using:
[
[H^+] = 10^{-\text{pH}}
]
[
[H^+] = 10^{-2.53} = 2.95 \times 10^{-3} \text{ M}
]
Step 3: Set up an ICE table
For the dissociation of the monoprotic acid ( HA ):
[
HA \rightleftharpoons H^+ + A^-
]
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| ( HA ) | 0.0192 | (-x) | (0.0192 – x) |
| ( H^+ ) | 0 | (+x) | (x) |
| ( A^- ) | 0 | (+x) | (x) |
Since ( [H^+] = x = 2.95 \times 10^{-3} ), we substitute this into the ICE table values.
Step 4: Apply the ( K_a ) expression
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
[
K_a = \frac{(2.95 \times 10^{-3})^2}{0.0192 – 2.95 \times 10^{-3}}
]
[
K_a = \frac{(8.70 \times 10^{-6})}{0.01625}
]
[
K_a = 5.36 \times 10^{-4}
]
Conclusion
The ( K_a ) value for the acid is ( 5.36 \times 10^{-4} ).
This result indicates the acid is weak because its ( K_a ) value is significantly less than 1. The method follows the assumption that ( x ) (the amount of dissociation) is small compared to the initial concentration, which is valid here.