Example 1
• Calculate the pH of the buffer solution
formed at 298 K when 0.125 mol of
sodium ethanoate is dissolved in 250.0
cm3 of a 1.00 mol dm3 solution of
ethanoic acid.
– [Ka of ethanoic acid is 1.70 x 10-5 mol dm-3
at 298 K.
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Volume = 250/1000 = 0.250
Sodium ethanoate, [Salt]
= 0.125 / 0.250 = 0.500
Ethanoic acid, [Acid] = 1.00
pH = -logKa + log [salt]/[acid]
= -log(1.70 x 10-5) + log(0.500/1.00)
= 4.47
Example 2
• Calculate the pH of a mixture (buffer
solution) formed by adding 25.0 cm3 of a
0.117 M (aq) solution of sodium
propanoate to 25.0 cm3 0.117 M (aq)
solution of propanoic acid.
• [Ka of propanoic acid, 1.35 x 10-5 M at 25
0C]
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Mol acid = 0.117 x (25/1000) = x
Mol salt = 0.117 x (25/1000) = x
Total V = 25 + 25 = 50 cm3 = 0.050 dm3
[Acid] = x/0.050 ; [salt] = x/0.050
Sub into formula…
pH = -logKa + log( [salt]/[acid])
= 4.87 answer