Buffer Calculation (1 Viewer)

[MS8]

a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100mL of this buffer

I got part a right but i don’t understand part b

 

[jimmysmith560]

Would the following working help?

Adding strong base will neutralise some of the acetic acid, yielding the conjugate base acetate ion. Compute the new concentrations of these two buffer components, then repeat the equilibrium calculation of part (a) using these new concentrations.



The initial molar amount of acetic acid is:



The amount of acetic acid remaining after some is neutralised by the added base is:



The newly formed acetate ion, along with the initially present acetate, gives a final acetate concentration of:



Compute molar concentrations for the two buffer components:



Using these concentrations, the pH of the solution may be computed as in part (a) above, yielding pH = 4.75 (only slightly different from that prior to adding the strong base).

 

[jazz519]

a) Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate.

b) Calculate the pH after 1.0 mL of 0.10 NaOH is added to 100mL of this buffer

I got part a right but i don’t understand part b

The answer above by jimmy is correct but it's not a logical way to do the question.

Use RICE tables. These are related to equilibrium equations you have already used in module 5/6 and so if you use that it's easy to set out the working logically and relate it to previous concepts you have learnt.

I'm assuming the question told you the Ka because you can't solve it without it. Acetic acid Ka = 1.8 x 10^(-5)

Solving a buffer question is actually very similar to if you were asked something like find the pH of a 0.1 mol/L acetic acid solution. The main difference is in the initial concentrations part of the RICE table


(a)
CH3COOH(aq) < -- > CH3COO-(aq) + H+(aq)

Ka = [CH3COO-] [H+] / [CH3COOH]

	CH3COOH	CH3COO-	H+
R	1	1	1
I	0.10	0.10	0
C	-x	+x	+x
E	0.10-x	0.10+x	x

This above RICE table is basically identical to what you do in the Ka questions but here it starts with CH3COO- being 0.10 instead of 0 normally because of the 0.10 M NaCH3COO solution.


1.8 x 10^-5 = (0.10+x)(x) / (0.10-x)
Small x approximation as K is small therefore, 0.10 + x = 0.10 and 0.10-x = 0.10

1.8 x 10^-5 = 0.10 x / 0.10
x = 1.8 x 10^-5

pH = -log[H+] = -log(1.8 x 10^-5) = 4.74472....


(b)
Do the same thing but instead of concentrations put moles in the RICE table

There's 100 mL buffer solution so
n(CH3COOH) = cv = (0.1)(0.1) = 0.01 moles
n(CH3COO-) = cv = (0.1)(0.1) = 0.01 moles

Theres 1 mL of NaOH so
n(NaOH) = cv = (0.1)(0.001) = 0.0001 moles

The NaOH will react with the CH3COOH and reduce the amount of it because acid + base reaction occurs (i.e. CH3COOH(aq) + NaOH(aq) --> CH3COONa(aq) + H2O(l)) and that forms CH3COONa (i.e. CH3COO- if you remove the Na+ spectator ion) and so CH3COO- amount goes up slightly

	CH3COOH	CH3COO-
R	1	1
I	0.01	0.01
C	-0.0001	+0.0001
E	0.0099	0.0101

Put these back into concentrations. The solution volume is 101 mL or 0.101 L


c(CH3COOH) = n/v = 0.0099/0.101 = 0.098... mol/L
c(CH3COO-) = n/v = 0.0101/0.101 = 0.10 mol/L

sub into Ka expression and you can basically just assume these concentrations are the same as at the equilibrium of dissociation step since we showed in part (a) the small x approximation applies

1.8 x 10^-5 = (0.10)[H+] / 0.098...
[H+] = 1.764... x 10^-5

pH = -log(1.764... x 10^-5) = 4.75