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What is the pH of 311.0 mL of a buffer that is 1.11MHF and 0.240MKF ? The Ka for HF is 3.50×10−4

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To find the pH of the buffer solution, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, HF is the acid (HA) and F- is the conjugate base (A-).

Given:

Volume of the buffer = 311.0 mL = 0.311 L

Concentration of HF (acid) = 1.11 M

Concentration of F- (conjugate base) = 0.240 M

Ka for HF = 3.50 × 10^(-4)

First, let's calculate the ratio of [A-]/[HA]:

[A-]/[HA] = (concentration of F-) / (concentration of HF)

[A-]/[HA] = 0.240 M / 1.11 M

[A-]/[HA] = 0.2162

Now, let's calculate the pKa:

pKa = -log(Ka)

pKa = -log(3.50 × 10^(-4))

pKa ≈ 3.4559

Finally, we can use the Henderson-Hasselbalch equation to find the pH:

pH = pKa + log([A-]/[HA])

pH = 3.4559 + log(0.2162)

pH ≈ 3.4559 - 0.6650

pH ≈ 2.7909

Therefore, the pH of the buffer solution is approximately 2.7909.

User Andrew Pate
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