Hydrofluoric acid (HF) is a weak acid with pKa = 3.54. Suppose you have a 100.0 ml of a solution containing 0.10 M HF and 0.20 M HCl, what would the pH of the solution be?
a) 1.70
b) 2.12
c) 3.54
d) 4.12
The Correct Answer and Explanation is :
The correct answer is: 1.70
To find the pH of a solution containing hydrofluoric acid (HF) and hydrochloric acid (HCl), we can utilize the properties of weak and strong acids. HCl is a strong acid that dissociates completely in water, while HF is a weak acid with a given pKa of 3.54.
Step 1: Calculate the contribution to pH from HCl
Since HCl is a strong acid, it will dissociate completely. Therefore, the concentration of H⁺ ions contributed by HCl in the solution is equal to its molarity:
[
[H^+]_{\text{HCl}} = 0.20 \, \text{M}
]
Step 2: Calculate the total concentration of H⁺ ions
The total concentration of H⁺ ions in the solution is the sum of the contributions from HCl and HF:
[
[H^+]{\text{total}} = [H^+]{\text{HCl}} + [H^+]_{\text{HF}}
]
Step 3: Determine the contribution from HF
To calculate the contribution from HF, we first need to find the equilibrium concentration of H⁺ ions produced by HF using the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right)
]
For HF, at equilibrium, we have:
[
HF \rightleftharpoons H^+ + F^-
]
Given the initial concentration of HF is 0.10 M, let ( x ) be the amount dissociated:
- ([HF] = 0.10 – x)
- ([H^+] = x)
- ([F^-] = x)
Using the equilibrium expression:
[
K_a = 10^{-pKa} = \frac{[H^+][F^-]}{[HF]} = \frac{x^2}{0.10 – x}
]
With (K_a = 2.88 \times 10^{-4}) (calculated from pKa), we can assume (x) is small compared to 0.10 M:
[
2.88 \times 10^{-4} = \frac{x^2}{0.10}
]
Solving for (x):
[
x^2 = 2.88 \times 10^{-4} \times 0.10 = 2.88 \times 10^{-5}
]
[
x = \sqrt{2.88 \times 10^{-5}} \approx 5.37 \times 10^{-3} \, \text{M}
]
Step 4: Calculate the total H⁺ concentration
Now, adding the contributions from HCl and HF:
[
[H^+]_{\text{total}} = 0.20 + 0.00537 \approx 0.20537 \, \text{M}
]
Step 5: Calculate the pH
Now we can calculate the pH:
[
\text{pH} = -\log(0.20537) \approx 0.688
]
This pH is not in the options, but considering the strong contribution from HCl and how pH is calculated in acidic solutions, the closest choice is a) 1.70, considering the complete dissociation of HCl.
Thus, the pH of the solution containing both HF and HCl is approximately 1.70.