Question

A 350.0 mL buffer solution is 0.170 M in HF and 0.170 M in NaF. A)...

A 350.0 mL buffer solution is 0.170 M in HF and 0.170 M in NaF.

A) What mass of NaOH could this buffer neutralize before the pH rise above 4.00?

B) If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?

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Homework Answers

Answer #1

A)

moles of HF = 350 x 0.170 / 1000 = 0.0595 mol

moles of NaF = 0.0595 mol

pH = pKa + log [salt + C / acid - C]

4.00 = 3.20 + log [0.0595 + C / 0.0595 - C]

[0.0595 + C / 0.0595 - C] = 6.3096

0.0595 + C = 0.375 - 6.3096 C

C = 0.0432

moles of NaOH = 0.0432 mol

mass of NaOH = 1.73 g

b)

moles of HF = NaF = 0.126

4.00 = 3.20 + log [0.126 + C / 0.126 - C]

[0.126 + C / 0.126 - C] = 6.3096

0.126 + C = 0.795 - 6.3096 C

C = 0.0915

moles of NaOH = 0.0915

mass of NaOH = 3.66 g

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