Question

Calculate the pH of a buffer, consisting of 0.42 M HF and 0.42 M F -, before and after addition of 0.44 g of NaOH to 1.0 L of the buffer (Ka of HF = 6.8 ✕ 10-4).

(a) before

(b) after

Answer 1

we know that

pKa = -log Ka = -log 6.8 x 10-4

pKa = 3.16749

for buffers

pH = pKa + log [conjugate base / acid]

in this case

pH = pKa + log [ F-/HF]

a) before adding NaOH

pH = 3.16749 + log[ 0.42 / 0.42]

pH = 3.16749 + log 1

pH = 3.16749 + 0

pH = 3.16749

so

pH of the buffer is 3.16749

b)

now

moles = mass / molar mass

moles of NaOH = 0.44 / 40 = 0.11

now

concentration = moles / volume

so

[NaOH] = 0.11 / 1 = 0.11

now

HF + OH- --> F- + H20

we can see that

[HF] reacted = [NaoH] added = 0.11

[F-] formed = [NaOH] added = 0.11

now finally

[HF] = 0.42 - 0.11 = 0.31

[F-] = 0.42 + 0.11 = 0.53

so

pH = pKa + log [F-/HF]

pH = 3.16749 + log [0.53 / 0.31]

pH = 3.4

so

the pH after addition of NaoH is 3.4