What is the pH of a aqueous buffer solution that is made of 0.237 M HF(aq) and 0.213 M KF(aq)?For HF, Ka = 3.548 x 10-4.Give your answer with 3 significant figures.
Question
What is the pH of a aqueous buffer solution that is made of 0.237 M HF(aq) and 0.213 M KF(aq)?For HF, Ka = 3.548 x 10-4.Give your answer with 3 significant figures.
Solution
To solve this problem, we will use the Henderson-Hasselbalch equation, which is used to estimate the pH of a buffer solution. The equation is as follows:
pH = pKa + log([A-]/[HA])
where:
- [A-] is the concentration of the base (F- from KF in this case)
- [HA] is the concentration of the acid (HF in this case)
- pKa is the negative log of the Ka value
First, we need to calculate the pKa from the given Ka.
pKa = -log(Ka) pKa = -log(3.548 x 10^-4) pKa ≈ 3.45
Then, we substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA]) pH = 3.45 + log(0.213/0.237) pH = 3.45 + log(0.8987) pH = 3.45 - 0.046 pH ≈ 3.40
So, the pH of the buffer solution is approximately 3.40.
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