mark scheme is WRONG???!!!
http://www.ocr.org.uk/Images/175443-question-paper-unit-f325-01-equilibria-energetics-and-elements.pdf
question 4dii)
[HCOOH] should be 64 imho.
the mark scheme says it's 24?!?
so {HCOOH} = 200/1000 * 3.2???
A chemist prepares a buffer solution by mixing together the following:200 cm of 3.20 mol dm–3 HCOOH (Ka= 1.70 × 10^-4mol dm-3) and800 cm3 of 0.500 mol dm-3 NaOH.The volume of the buffer solution is 1.00 dm3. • Explain why a buffer solution is formed when these two solutions are mixed together. • Calculate the pH of this buffer solution.
question 4dii)
[HCOOH] should be 64 imho.
the mark scheme says it's 24?!?
so {HCOOH} = 200/1000 * 3.2???
A chemist prepares a buffer solution by mixing together the following:200 cm of 3.20 mol dm–3 HCOOH (Ka= 1.70 × 10^-4mol dm-3) and800 cm3 of 0.500 mol dm-3 NaOH.The volume of the buffer solution is 1.00 dm3. • Explain why a buffer solution is formed when these two solutions are mixed together. • Calculate the pH of this buffer solution.
n(HCOOH) = c.v = 3.20 x 200/1000 = 0.640 mol
n(NaOH) = 0.500 x 800/1000 = 0.400 mol
n(HCOOH) remaining, after being partially neutralised by the NaOH = 0.640 - 0.400 = 0.240 mol
Since combined volume to 1 dm3, [HCOOH] after partial neutralisation = 0.240/1 = I'll let you work that one out
Recent marks schemes should never be wrong, or at least once they get to being released.
n(NaOH) = 0.500 x 800/1000 = 0.400 mol
n(HCOOH) remaining, after being partially neutralised by the NaOH = 0.640 - 0.400 = 0.240 mol
Since combined volume to 1 dm3, [HCOOH] after partial neutralisation = 0.240/1 = I'll let you work that one out
Recent marks schemes should never be wrong, or at least once they get to being released.
what about the NaOH? why does it suddenly change to HCOONa? Also, why are the moles for HCOO- and HCOONa the same?
Original post by innieminnieminy
what about the NaOH? why does it suddenly change to HCOONa? Also, why are the moles for HCOO- and HCOONa the same?
There is an excess of the acid in the mixture. All of the NaOH reacts with the acid giving HCOONa.
Hence, the mol of NaOH initially become the same mol of HCOONa = 0.8 x 0.5 = 0.40 mol
The mol of acid remaining is the initial mol - the reacted mol = (0.2 x 3.2) - 0.4 = 0.24 mol
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