you have a 250 ml sample of 1.00 M acetic acid (ka= 1.8x10^-5)?

Question

Assuming no volume change, how much NaOH must be added to make the best buffer?
A.10.0g
B. 5.0g
C. 15.0g
D. 20.0g
E. none of these

please provide explanation
i cant figure out buffers for myl ife

Answer 1

The best buffer is always when [acid]=[conjugate base]
NOTE that these are equilibrium and not total/initial concentrations.

in your case 1.00 M CH3COOH is the inital acid concentration and also it will be the total acetate concentration and therefore the concentration of the buffer.
The conjugate base is CH3COO-
You need [CH3COOH] = [CH3COO-] (1)
but Ctotal= [CH3COOH] +[CH3COO-] =1.00 M (2)
so from (2) substituting with (1)
2*[CH3COO-]=1 => [CH3COO-]=0.50 M

This will come from the reaction of acetic acid with NaOH.
Since the stoichiometry for all is 1:1, that means that you will need
C(NaOH)=0.5 M
mole=mass/MW and also
mole= C*V

so mass/MW=M*V =>
mass = M*V*MW= 0.5*0.25*40 = 5.0 g

so the answer is B

Answer 2

You have 0.25 moles of ethanoic acid present. The best buffer will be one which has equal numbers of moles of ethanoic acid and sodium ethanoate. So you need to neutralise half of the ethanoic acid, and this will require 0.25/2 moles. This corresponds to 5g of sodium hydroxide.