Question

A 360.0 −mL buffer solution is 0.160 M in HFand 0.160 M in NaF.

Part A)

What mass of NaOH could this buffer neutralize before the pH rises above 4.00?

Part B)

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

Answer 1

Ka of HF = 7.2 x 10^-4

pKa = - log Ka = 3.14

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

4.00 = 3.14 + log [F-]/ [HF]

4.00 - 3.14 = log [F-]/ [HF]

0.86 = log [F-]/ [HF]

10^0.86 = log [F-]/ [HF]=7.24

7.24 = (0.160+x) / (0.160-x)

1.16 - 7.24 x = 0.160 + x

1 = 8.24 x

x = 0.121

moles NaOH = 0.121 x 0.340 L= 0.041 mol

mass NaOH = 0.041 x 40 g/mol = 1.64 g

(b)

Ka of HF = 7.2 x 10^-4

pKa = - log Ka = 3.14

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

4.00 = 3.14 + log [F-]/ [HF]

4.00 - 3.14 = log [F-]/ [HF]

0.86 = log [F-]/ [HF]

10^0.86 = log [F-]/ [HF]=7.24

7.24 = (0.350+x) / (0.350-x)

2.534 - 7.24 x = 0.350 + x

2.184 = 8.24 x

x = 0.265

moles NaOH = 0.265 x 0.340 L= 0.090 mol

mass NaOH = 0.090 x 40 g/mol = 3.60 g