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# pH/Buffer questions watch

1. Hi, may somebody help me understand how to approach these questions. (questions d-f) I can do a-c fine. I have the answers with me, but don't know how to work them out. I'll put the answers in bol and my calculations in a spoiler.

a) What is meant by the term "buffer solution"?

b) Calculate the pH of a buffer solution which contains the weak monoprotic acid, propanoic acid (CH3CH2COOH), in concentration 0.1 moldm^-3 and sodium propanoate in concentration 0.05 moldm^-3. Ka of propanoic acid is 1.26 x 10^-5 moldm^-3.
4.60

Spoiler:
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[H+] = Ka * [acid] / [base]
[H+] = 1.26 x 10^-5 * 0.1/0.05
[H+] = 2.52 x 10^-5
pH = -log10(2.52 x 10^-5)
pH = 4.60

c) Give equations to show how the above solution fulfills its buffer function.

d) Calculate the pH of the solution after 0.01 moles of NaOH are added to 500 cm3 of the solution.
4.84
e) Calculate the pH of the solution after 0.01 moles of HCl are added to 500 cm3 of the solution.
4.30
f) Calculate the pH after 0.01 moles of NaOH is added to 500 cm3 of water.
12.30
2. (Original post by purplefrog)
Hi, may somebody help me understand how to approach these questions. (questions d-f) I can do a-c fine. I have the answers with me, but don't know how to work them out. I'll put the answers in bol and my calculations in a spoiler.
here is everything you need

It really is a great explanation. And perfect for A level
3. (Original post by Plato's Trousers)
here is everything you need

It really is a great explanation. And perfect for A level
chemguide addresses how to questions like part C), but I can't seem to see anything on how to tackle problems posed by d-f :s
4. (Original post by purplefrog)
chemguide addresses how to questions like part C), but I can't seem to see anything on how to tackle problems posed by d-f :s
These are stoichiometry followed by re-calculation of concentrations before using the buffer law once again.

For example in d) you are adding NaOH solution.

Calculate the moles of NaOH added.
Using the reaction equation, calculate the moles of acid used up and hence left over, and salt formed.
Using the new volume calculate the new concentrations.
Apply buffer law...
5. 0.01 moles of NaOH are added to 500 cm3 of the solution.

(Original post by charco)
Calculate the moles of NaOH added.
0.01

Using the reaction equation, calculate the moles of acid used up
Ka = 1.26 x 10^-5
Ka = [H+]^2/0.1
0.1 * Ka = [H+]^2
[H+] = 1.26 x 10^-6 mol dm^-3
500cm^3 = 0.5dm^3
0.5 [H+] = 6.3 x 10^-7

6.3 x 10^-7 moles of acid present in 500 cm3 of solution.

all moles of acid used up: 0.01 - 6.3 x 10^-7 (moles of OH - moles of acid)

leaves 9.99 x 10^-3 moles of OH left.

where do I go from here?

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