Buffer calculations

Kahynickel SG

A solution containing both Na2HPO4 and NaH2PO4 is commonly used as a buffer solution. The following equilibrium is present in the solution

H2PO4–(aq) = HPO42–(aq) + H+(aq) Ka = 6.2 × 10–8 mol dm–3

Calculate the pH of a buffer solution made by mixing 100 cm3 of 0.5 mol dm–3 Na2HPO4 and 100 cm3 of 0.3 mol dm–3 NaH2PO4.

Solution: We need to have the concentartions of both H2PO4- and HPO42 in oreder to calculate the pH. The point of confusion is the CIE marking scheme uses the concentrations given in the questions i.e [Na2HPO4]= 0.5 M and [NaH2PO4]=0.3 M and put it in the Handerson-Hasselbach equation.

The total volume of the mixture is now 100 + 100 = 200 cm3.

so the new concentrations must be calculated using this new volume.

[Na2HPO4] = 0.3 x 100/200 = 0.15 M

and then [NaH2PO4]

Please explain.

Marn shun1 SG

So basically they are talking about how the concentrations of the salt and weak acid given in the question will actually change their concentrations due to the new total volume when both solutions are mixed together. Therefore in order to find the pH of the buffer solution, we must use the Handerson-Hasselbach equation using the newly found concentration of the weak acid and its salt, which is derived from dividing the amount of salt/acid (mols) with the total volume of the solution which is 200cm^3 in this case. The calculations for the Na2HPO4 is wrong, you should get 0.25 mol dm-3 for [NA2HPO4].

Keep in mind you need to perform your calculationsin terms of dm3 for the volumes

Kahynickel SG

yes that would be 0.25M. But are my calculations correct?

UltimaOnline SG

2 points to note :

1stly, for Cambridge A levels, as long as your final answer is correct, you'll get full marks, regardless of whether your working is the same as that given in the mark scheme (unless your working is obviously based on erroneous concepts or assumptions).

2ndly, for this question, because Cambridge (too kindly) used the same volumes for both members of the conjugate acid-base pair, hence upon dilution (ie. mixing both volumes together), the new ratio of molarities will still be the same as the original ratio of molarities.

Hence, Cambridge simply used the original molarities, plugged into the Henderson-Hasselbalch equation, instead of calculating the new molarities (as you've correctly done, no worries). Final answer will still be the same.

Kahynickel SG

okay thank you.

Another query from CIE Nov 2013 paper.

Buffer solution F was prepared by adding 0.0300 mol of sodium hydroxide to 100 cm3 of a 0.500 mol dm–3 solution of propanoic acid.

Calculate the concentrations of propanoic acid and sodium propanoate in buffer solution F.

[propanoic acid] = ................................ mol dm–3

[sodium propanoate] = ................................ mol dm–3

Solution:

no of moles of acid= 0.05

remaining moles = 0.03

moles of sodium propanoate = moles of sodium hydroxide = 0.03

Now these are moles we need to convert them to concentrations but the volume of NaOH is not given. so these are taken as concentration. Is it correct?

UltimaOnline SG

Yes, correct.

Do an ICF (not ICE) table in moles.

Final moles of propanoic acid = 0.05 - 0.03 = 0.02 mol

Final moles of propanoate ion = 0 + 0.03 = 0.03 mol

Final molarity of propanoic acid = (0.02 / (100/1000)) = 0.2 M

Final molarity of propanoate ion = (0.03 / (100/1000)) = 0.3 M

To calculate pH, you can plug either the ratio of molarities, or simply the ratio of moles (since the same volume in both numerator and denominator will cancel out), of [conjugate base] / [conjugate acid] into the Henderson-Hasselbalch equation.

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