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 :
Solution:
To calculate the acid dissociation constant (( K_a )) for the monoprotic acid, we use the given information:
- Concentration of the acid: ( [HA] = 0.0192 ) M
- pH of the solution: ( 2.53 )
Step 1: Determine ([H^+])
The concentration of hydrogen ions (([H^+])) is found using the pH equation:
[
pH = -\log [H^+]
]
[
[H^+] = 10^{-pH} = 10^{-2.53}
]
[
[H^+] = 2.95 \times 10^{-3} \text{ M}
]
Step 2: Set up the ICE Table for Dissociation
A monoprotic acid (( HA )) dissociates as:
[
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^+]) at equilibrium is ( 2.95 \times 10^{-3} ) M, we set ( x = 2.95 \times 10^{-3} ).
[
[HA]_{eq} = 0.0192 – 2.95 \times 10^{-3} = 0.01625 \text{ M}
]
Step 3: Calculate (K_a)
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
[
K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625}
]
[
K_a = \frac{8.70 \times 10^{-6}}{0.01625}
]
[
K_a = 5.36 \times 10^{-4}
]
Final Answer:
[
K_a = 5.36 \times 10^{-4}
]
Explanation (300 Words)
The acid dissociation constant (( K_a )) measures the strength of an acid in aqueous solution. A higher ( K_a ) value means a stronger acid, while a lower ( K_a ) value indicates a weaker acid. To determine ( K_a ), we must first calculate the equilibrium concentration of hydrogen ions (([H^+])) using the given pH.
The pH is related to ([H^+]) by the equation:
[
[H^+] = 10^{-pH}
]
Using ( pH = 2.53 ), we calculate ([H^+] = 2.95 \times 10^{-3} ) M. Since the acid is monoprotic (( HA )), it dissociates into ( H^+ ) and ( A^- ) in a 1:1 ratio. This means the concentration of ( A^- ) at equilibrium is also ( 2.95 \times 10^{-3} ) M.
The initial acid concentration is given as 0.0192 M. At equilibrium, the remaining ( HA ) concentration is found by subtracting the amount dissociated:
[
[HA]_{eq} = 0.0192 – 2.95 \times 10^{-3} = 0.01625 \text{ M}
]
Using the expression for ( K_a ):
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substituting values, we find:
[
K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} = 5.36 \times 10^{-4}
]
Thus, the acid has a dissociation constant of ( 5.36 \times 10^{-4} ), indicating a weak acid.