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 use the relationship between the acid concentration, pH, and Ka. Here’s how we can approach this problem step-by-step:
Step 1: Determine the concentration of H⁺ ions from the pH
The pH is given as 2.53. We can calculate the concentration of hydrogen ions (H⁺) using the following formula:
$$
\text{pH} = -\log[\text{H}^+]
$$
Rearranging to solve for $[\text{H}^+]$:
$$
[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
$$
So, the concentration of hydrogen ions is approximately $2.95 \times 10^{-3} \, \text{M}$.
Step 2: Set up the dissociation equation for the acid
For a monoprotic acid HA, the dissociation in water is represented as:
$$
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
$$
Initially, before dissociation, the concentration of HA is 0.0192 M, and the concentrations of H⁺ and A⁻ are both 0 M. After dissociation, let the amount of HA dissociate be $x$. At equilibrium:
- $[HA] = 0.0192 – x$
- $[H^+] = x$
- $[A^-] = x$
Since we know the concentration of $H^+$ at equilibrium is $2.95 \times 10^{-3} \, \text{M}$, we set $x = 2.95 \times 10^{-3} \, \text{M}$.
Step 3: Apply the equilibrium expression for Ka
The acid dissociation constant $K_a$ is given by:
$$
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
$$
Substitute the known values:
$$
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.0192 – 2.95 \times 10^{-3}} = \frac{8.70 \times 10^{-6}}{0.01625}
$$
$$
K_a \approx 5.36 \times 10^{-4}
$$
Final Answer:
The $K_a$ for the acid is approximately $5.36 \times 10^{-4}$.
Explanation:
This problem involves understanding how a monoprotic acid dissociates in water and how the pH relates to the concentration of hydrogen ions. The concentration of hydrogen ions, calculated from the pH, allows us to determine the extent of dissociation, which is essential for calculating the acid dissociation constant $K_a$. By setting up the equilibrium expression for the dissociation reaction, we can substitute the known values to solve for $K_a$, which tells us about the strength of the acid in water. The smaller the $K_a$, the weaker the acid, as it dissociates less in solution.