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
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
Get Answers For Free
Most questions answered within 1 hours.