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 can use the information provided: the concentration of the acid solution (0.0192 M) and the pH of the solution (2.53).
Step 1: Determine the concentration of ([H^+])
From the pH of the solution, we can calculate the concentration of hydrogen ions ([H^+]) using the following formula:
[
\text{pH} = -\log [H^+]
]
Given that the pH is 2.53, we can rearrange the equation to solve for ([H^+]):
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
]
So, ([H^+] = 2.95 \times 10^{-3} \, \text{M}).
Step 2: Set up an ICE table
We will use an ICE (Initial, Change, Equilibrium) table to understand how the dissociation of the monoprotic acid proceeds in water. The dissociation of a monoprotic acid, HA, in water can be represented as:
[
\text{HA} (aq) \rightleftharpoons \text{H}^+ (aq) + \text{A}^- (aq)
]
Initial concentrations:
- ([HA]) = 0.0192 M (given)
- ([H^+]) = 0 M
- ([A^-]) = 0 M
Change in concentrations:
Since each mole of HA dissociates to produce one mole of (H^+) and one mole of (A^-), if x is the amount of HA that dissociates, the change in concentrations will be:
- ([HA]) decreases by x
- ([H^+]) increases by x
- ([A^-]) increases by x
Equilibrium concentrations:
- ([HA]) = 0.0192 – x
- ([H^+]) = x
- ([A^-]) = x
From the pH calculation, we know that ([H^+] = 2.95 \times 10^{-3} \, \text{M}), so (x = 2.95 \times 10^{-3}\, \text{M}).
Step 3: Solve for Ka
The expression for the acid dissociation constant (K_a) is:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substitute the equilibrium concentrations into this expression:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}}
]
First, calculate the numerator:
[
(2.95 \times 10^{-3})^2 = 8.70 \times 10^{-6}
]
Now calculate the denominator:
[
0.0192 – 2.95 \times 10^{-3} = 0.01625 \, \text{M}
]
Now, calculate (K_a):
[
K_a = \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.35 \times 10^{-4}
]
Final Answer:
The acid dissociation constant (K_a) for the acid is approximately:
[
K_a \approx 5.35 \times 10^{-4}
]
Explanation:
The acid dissociation constant (K_a) is a measure of the strength of an acid in solution. A larger (K_a) indicates a stronger acid, meaning it dissociates more completely in water. In this case, the value of (K_a = 5.35 \times 10^{-4}) suggests that the acid is a weak acid, as the dissociation is not complete, and only a small fraction of the acid dissociates to produce (H^+) ions. By using the pH to find the concentration of hydrogen ions and applying it to the equilibrium expression, we can calculate the (K_a), which gives us insight into the acid’s dissociation behavior.