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 determine the acid dissociation constant (( K_a )) of the monoprotic acid, we follow these steps:
Step 1: Define Known Values
- Concentration of acid: ( [HA] = 0.0192 ) M
- pH of solution: ( 2.53 )
Step 2: Calculate the ( [H^+] ) Concentration
The hydrogen ion concentration can be determined from the pH formula:
[
[H^+] = 10^{-\text{pH}}
]
[
[H^+] = 10^{-2.53}
]
[
[H^+] = 2.95 \times 10^{-3} \text{ M}
]
Step 3: Set Up the ICE Table
For the dissociation of a 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} ) M, we substitute this into the equilibrium expression.
Step 4: Write the ( K_a ) Expression
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
[
K_a = \frac{(2.95 \times 10^{-3}) (2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}}
]
Step 5: Solve for ( K_a )
[
K_a = \frac{(8.70 \times 10^{-6})}{0.0163}
]
[
K_a = 5.34 \times 10^{-4}
]
Final Answer:
[
K_a = \mathbf{5.34 \times 10^{-4}}
]
Explanation (300 Words)
The acid dissociation constant (( K_a )) quantifies the strength of a weak acid in solution. A higher ( K_a ) value indicates stronger acid dissociation, meaning more ( H^+ ) ions are produced.
To calculate ( K_a ), we start by determining the hydrogen ion concentration using the given pH. Since ( pH = -\log [H^+] ), we take the inverse log (antilog) to find ( [H^+] ). The result, ( 2.95 \times 10^{-3} ) M, represents the equilibrium concentration of both ( H^+ ) and ( A^- ) because they are produced in a 1:1 ratio from the acid.
Next, we construct an ICE table (Initial, Change, Equilibrium) to track how the acid dissociates. Initially, the acid concentration is 0.0192 M, and since some dissociates into ( H^+ ) and ( A^- ), we subtract ( x ) (which equals ( [H^+] )) from the initial concentration of ( HA ). Using these equilibrium values, we apply the expression for ( K_a ), substitute the known values, and solve for ( K_a ).
The final answer, ( 5.34 \times 10^{-4} ), confirms that the acid is weak (since ( K_a ) values for strong acids are typically much larger). This method is crucial in chemistry for determining acid strength, buffer capacities, and pH behavior in various solutions.