# Chemistry a level buffers questions

To prepare a buffer solution, 75.0 cm3 of 0.220 mol dm–3 butanoic acid is reacted with 50.0 cm3 of 0.185 mol dm–3 sodium hydroxide.

Ka for butanoic acid is 1.5 × 10–5 mol dm–3 at 25 °C

Calculate the pH of the buffer solution at 25 °C. Give your answer to two decimal places.

can you please show your working out as im confused to how you calculate this.
Work out mol of acid and base:
For Butanoic acid you do (75/1000) x 0.22 = 0.0165mol
For NaOH you do (50/1000) x 0.185 = 9.25x10-3 mol

Work out mol remaining after the reaction:
0.0165 - 9.25x10-3 = 7.25x10-3 mol

Use -log([H+]) to find pH:
-log(7.25x10-3) = 2.14
To prepare a buffer solution, 75.0 cm3 of 0.220 mol dm–3 butanoic acid is reacted with 50.0 cm3 of 0.185 mol dm–3 sodium hydroxide.

Ka for butanoic acid is 1.5 × 10–5 mol dm–3 at 25 °C

Calculate the pH of the buffer solution at 25 °C. Give your answer to two decimal places.

can you please show your working out as im confused to how you calculate this.

Using the Henderson-Hasselbach equation:

pH buffer = pKa + log([Salt]÷[Acid])
Note: [] means concentration
pKa = -log(Ka)
pKa = -log (1.5 x 10^-5)
pKa = 4.824

The number of moles of hydroxide is limiting/isn't in excess, so its moles is the minimum number of moles of salt that can be formed, as there isn't enough hydroxide left over to react with the excess acid.

n(NaOH) = 50/1000 x 0.185 = 9.25 x 10^-3
1 mole NaOH = 1 mole CH3CH2COONa
So moles of salt is 9.25 x 10^-3
[Salt] = 9.25 x 10^-3/ 125/1000 = 0.074 mol dm-3
[Acid] = 7.25 x 10^-3/ 125/1000 = 0.058

Plugging this back into the equation gives
pH buffer = 4.824 + log(0.074/0.058)

pH = 4.93
(edited 6 years ago)
Original post by Hajra Momoniat
Work out mol of acid and base:
For Butanoic acid you do (75/1000) x 0.22 = 0.0165mol
For NaOH you do (50/1000) x 0.185 = 9.25x10-3 mol

Work out mol remaining after the reaction:
0.0165 - 9.25x10-3 = 7.25x10-3 mol

Use -log([H+]) to find pH:
-log(7.25x10-3) = 2.14

You havent taken the new total volume into account nor used the Ka value given
Actually its pH=4.93
Ph = pka log[A-]/[HA]
Ph = -log(1.5×10^-3) log(0.185/0.220)
Put it in calculator
Ph = 2.74
(edited 4 years ago)
Original post by Mansdeep
Ph = pka log[A-]/[HA]
Ph = -log(1.5×10^-3) log(0.185/0.220)
Put it in calculator
Ph = 2.74

Your calculation assumes that you're adding butanoate ions to butanoic acid. In this case, NaOH is added, which partially neutralises the butanoic acid, generating butanoate ions.

If you re-read the thread, you'll find the values you should have used to get 4.93, which is the correct answer.
Original post by Hajra Momoniat
Work out mol of acid and base:
For Butanoic acid you do (75/1000) x 0.22 = 0.0165mol
For NaOH you do (50/1000) x 0.185 = 9.25x10-3 mol

Work out mol remaining after the reaction:
0.0165 - 9.25x10-3 = 7.25x10-3 mol

Use -log([H+]) to find pH:
-log(7.25x10-3) = 2.14

no