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A buffer was made by mixing 0.1070 moles of HF with .1147 moles of NaF and...

A buffer was made by mixing 0.1070 moles of HF with .1147 moles of NaF and diluting to exactly 1 liter. What will be the pH after addition of 10.00 mL of 0.2048 M HCl to 50.00 mL of the buffer? Ka(HF) = 3.500e-4. Note: only a portion of the original buffer is used in the second part of the problem. Answer is 3.154

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Answer #1

Volume of the buffer solution = 1 L

Given:ka (HF) = 3.50 x 10-4

pKa = -log(3.50 x 10-4) = -log3.50 + 4log10 = 3.45

Using Henderson - Hasselbalch equation:

pH = pKa + log([F-]/[HF])

[F-] = no. of moles / Volume of solution = 0.1070 / 1 = 0.107M

[NaF] = 0.1147 M

pH = 3.45 + log(0.1147/0.107) = 3.48

Now After HCl is added to 50 ml of this buffer

HCl + NaF HF + NaCl

Concentration of HF increases and of NaF decreases.

Number of moles of HCl added = volume x molarity = 0.01 L x 0.2048 = 0.002048 moles

Moles of NaF after addition of HCl = 0.005735 - 0.002048 = 0.003687

New [NaF] = 0.003687 / 0.06 = 0.06145 M

Moles of HF after HCl addition = 0.00535 + 0.002048 = 0.007398

New [HF] = 0.007398 / 0.06 = 0.1233 M

pH = pKa + log([F-]/[HF]) = 3.45 + log (0.06145 /0.1233) = 3.147 (after addition of HCl)

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