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Chemistry a level buffers questions
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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.
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.
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#2
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
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
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#3
(Original post by AnonymousAnswer)
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.
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.
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
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#4
(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
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
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#6
Ph = pka log[A-]/[HA]
Ph = -log(1.5×10^-3) log(0.185/0.220)
Put it in calculator
Ph = 2.74
Ph = -log(1.5×10^-3) log(0.185/0.220)
Put it in calculator
Ph = 2.74
Last edited by Mansdeep; 5 months ago
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#7
(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
Ph = pka log[A-]/[HA]
Ph = -log(1.5×10^-3) log(0.185/0.220)
Put it in calculator
Ph = 2.74
If you re-read the thread, you'll find the values you should have used to get 4.93, which is the correct answer.
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