Buffer solutions

Hi could someone help me with questions 3,4,5 with the working out because I don’t get any of these questions thank you BED59968-9CEA-4400-8843-FE38CB2C21BD.jpg.jpeg

BED59968-9CEA-4400-8843-FE38CB2C21BD.jpg.jpeg

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Reply 1

ASOC01

Hi, I don't think my exam board (OCR) includes weak base buffer solution, so I'm not sure if I got question 3 right as I just looked it up on chemguide and learn about it. And sorry for the messy structure as I didn't realise question 5 contains 5 separate questions itself! Also not sure about 5e, I just read it somewhere in another thread. Hope you can understand my working out! :smile:

P.S. ignore the vectors calculation in the mid section lol sorry forgot to cross it out

Original post by Puja_xo

Hi could someone help me with questions 3,4,5 with the working out because I don’t get any of these questions thank you BED59968-9CEA-4400-8843-FE38CB2C21BD.jpg.jpeg

These are simply a matter of applying the weak acid equilibrium equation or the Henderson/Hasselbalch equation which is derived from the former.

Ka = [H+][A-]/[HA]

where [H+] is the concentration of hydrogen ions from the acid, and is given by the pH = -log₁₀[H+]
[A-] is the concentration of the conjugate base of the acid and is equal to the salt concentration
[HA] is the weak acid concentration

In question 5 you first have to do some stoichiometry to find the final concentrations of the components. These are acid base reactions and all have stoichiometry 1:1, so it is fairly simple.

For example in 5c you have to calculate the mol of salt formed when 0.05 mol of NaOH added, and also the mol of acid left over = initial mol acid - mol reacted

Reply 3

Puja_xo

Could you explain the numbers that you’ve written under the equations for 5b,5c,5d please (the 0.05,08)
I checked the answers and all were correct thank you apart from 3 which they got 8.65

Reply 4

ASOC01

Original post by Puja_xo

Could you explain the numbers that you’ve written under the equations for 5b,5c,5d please (the 0.05,08)
I checked the answers and all were correct thank you apart from 3 which they got 8.65

oops, sorry abt q3 hahaha I shouldn't have written it down as I wasn't taught abt it. Which exam board is this if you don't mind me asking?
For the numbers in q5b,c, and d - they are referring to the amount of moles of each substances in the solution. It's the same number as the concentration because the question stated that there is 1dm^3 of solution. :smile:

Reply 5

charco

Study Forum Helper

Original post by Puja_xo

Could you explain the numbers that you’ve written under the equations for 5b,5c,5d please (the 0.05,08)
I checked the answers and all were correct thank you apart from 3 which they got 8.65

Ka = [H+][A-]/[HA]

Q3. pH of a buffer which is 0.2 mol dm^-3 in ammonium sulphate and 0.1 mol dm^-3 in ammonia.

ammonium sulfate has the formula (NH₄)₂SO₄ and hence a 0.2M solution is actually 0.4 M in ammonium ions.

equation for the a/dissociation

NH₄⁺ <==> NH₃ + H⁺

5.6 x 10^-10 = [H+]*0.1/0.4

[H+] = 4 * 5.6 x 10^-10

[H+] = 2.24 x 10^-9

pH = 8.65

Reply 6

ASOC01