Question

A buffer is created by dissolving 0.450 mol of dodium ascorbate and 0.500 mol ascorbic acid in enough water to create 1.00L of solution. The Ka of ascorbic acid is 7.9 x 10^-5.

Answer 1

You can use the henderson-hasselbach equation: pH = pKa + log\(\sf \frac{[A^-]}{[HA]}\)

Answer 2

Make sure that the mass and volume added does not surpass the titration point, the volume at which pH is at equilibrium.

Answer 3

A)PH=Pka+log([A-]/[HA])
PH=-log (7.9x10^-5)+log (0.45/0.50)
PH=4.10-0.0458
PH=4.05

B) When you add 0.015mol HCl the number of moles of the acid will increase and the number of moles of the base will decrase.
new number of moles of acid = 0.5+0.015=0.515mol
new number of mole of base = 0.45-0.015=0.435mol
Pka=-log[Ka]=-log[7.9*10^-5]=4.102

PH=Pka+log[Number of moles of base/number of moles of acid]
pH=4.102+log[0.435/0.515]=4.102+0.073=4.03

C) Mw of NaOH = 40g/mol
Number of moles = mass/Mw = 1/40=0.025moles NaOH ( Base ) Added
new number of moles of base = 0.025+ 0.45 =0.475mol
new number of moles of acid = 0.5-0.025 = 0.475mol
pH = 4.102+log[1] = 4.102

is this right?

Answer 4

Yes, you need to calculate the amount of moles going in and going out. You will need to do some math.

Answer 5

Yes, I am not near a calculator, but It looks correct from what I am looking at here.

Answer 6

I'm not sure if i need to write the balanced equation, the Ka=/ and the ICE chart or not

Answer 7

I suppose if it helps you see things more clear, you can make an "ICE" table.