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Buffers questions help please

A student adds 50.0cm^3 of 0.250 mol dm^-3 butanoic acid to 50.0cm^3 of 0.0500 mol dm^-3 sodium hydroxide. A buffer solution forms.

i) Explain why a buffer solution forms

ii) Calculate the pH of the buffer solution.

The Ka of butanoic acid is 1.51 x a0^-5 mold dm^-3

Our teacher didnt teach us this method on making a buffer only the other one by adding a weak acid to its conjugate base
Original post by IShouldBeRevising_
A student adds 50.0cm^3 of 0.250 mol dm^-3 butanoic acid to 50.0cm^3 of 0.0500 mol dm^-3 sodium hydroxide. A buffer solution forms.

i) Explain why a buffer solution forms

ii) Calculate the pH of the buffer solution.

The Ka of butanoic acid is 1.51 x a0^-5 mold dm^-3

Our teacher didnt teach us this method on making a buffer only the other one by adding a weak acid to its conjugate base


i) after the reaction between the acid and the base you are left with only the weak acid and its salt. So in this case you must do the stoichiometry calculation first to find the moles of acid remaining and the moles of salt formed.

You then use ka = [H+][A-]/[HA] as normal.
Original post by charco
i) after the reaction between the acid and the base you are left with only the weak acid and its salt. So in this case you must do the stoichiometry calculation first to find the moles of acid remaining and the moles of salt formed.

You then use ka = [H+][A-]/[HA] as normal.


All the NaOh reacts so your left with

weak acid --> salt of weak acid+ H20 ?
Original post by IShouldBeRevising_
All the NaOh reacts so your left with

weak acid --> salt of weak acid+ H20 ?


... and also the excess acid (very important)
Original post by charco
... and also the excess acid (very important)


yeah.. if H+ was added it would react with the salt of the weak acid and produce the weak acid oh- added it would react with the weak acid. Would the salt ion just be floating about
Not sure if you still need help?

Forms a buffer as people have explained; HA + OH- --> A- + H2O

Calculation.

Moles of OH- = M x V / 1000 = 2.5 x 10^-3
Moles of A- thus equals 2.5 x 10^-3 (as shown in the initial equation)
Moles of HA = M x V / 1000 = 0.0125

Buffer Concentrations
Conc HA = 1000/TOTAL VOLUME x 0.0125 = 1000/100 x 0.0125 = 0.125mol dm^-3
Conc A- = 1000/100 x 2.5 x 10^-3 = 0.025

Rearrange Ka..

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

Sub in.

[H+] = (1.51 x 10^-5) x (0.125) / (0.025)
=7.55 x 10^-5
-log10[H+]

ph = 4.12

-- i think--
Original post by Me-A-Doctor?

Buffer Concentrations
Conc HA = 1000/TOTAL VOLUME x 0.0125 = 1000/100 x 0.0125 = 0.125mol dm^-3
Conc A- = 1000/100 x 2.5 x 10^-3 = 0.025




Why is it you do 1000/Total volume ?
The way I calculated it was:

c = n / v
c = concentration
n = number of moles
v = volume in cm3

[HA] = 0.0125 / (100/1000) = 0.125 mol dm^-3
[A-] = 2.5 x 10^-3 / (100/1000) = 0.025 mol dm^-3

I divided the 100cm3 by 1000 to get a value in dm3
Original post by Slayingtheundead
The way I calculated it was:

c = n / v
c = concentration
n = number of moles
v = volume in cm3

[HA] = 0.0125 / (100/1000) = 0.125 mol dm^-3
[A-] = 2.5 x 10^-3 / (100/1000) = 0.025 mol dm^-3

I divided the 100cm3 by 1000 to get a value in dm3


I imagine that the question poser has been on tender-hooks for the past 2 years and 11 months
Reply 9
Original post by Me-A-Doctor?
Not sure if you still need help?

Forms a buffer as people have explained; HA + OH- --> A- + H2O

Calculation.

Moles of OH- = M x V / 1000 = 2.5 x 10^-3
Moles of A- thus equals 2.5 x 10^-3 (as shown in the initial equation)
Moles of HA = M x V / 1000 = 0.0125

Buffer Concentrations
Conc HA = 1000/TOTAL VOLUME x 0.0125 = 1000/100 x 0.0125 = 0.125mol dm^-3
Conc A- = 1000/100 x 2.5 x 10^-3 = 0.025

Rearrange Ka..

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

Sub in.

[H+] = (1.51 x 10^-5) x (0.125) / (0.025)
=7.55 x 10^-5
-log10[H+]

ph = 4.12

-- i think--

Thanks 👍🏻

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