Buffer question - very hard, 15points

a) pH = -log( [H+] )
5 = -log [H+]
[H+] = 10-5 M

b) pH = pKa + log( [A-]/[HA] )
5 = -log(1.8X10-5) + log( [A-]/[HA] )
5 - 4.74 = 0.26 = log( [A-]/[HA] )
[A-]/[HA] = 10^0.26 = 1.30
[acetic acid]/[acetate] = 1/1.30 = 0.769

c) Clearly, it is 1.3 mole of NaC2H3O2

2006-11-03 20:52:38 補充：
b) pH = pKa + log( [A-]/[HA] )

2006-11-03 20:53:53 補充：
I do not know why it becomes ***** .It is -log(Ka)

You have been asked to prepare a buffer solution with pH=5, using sodium acetate and acetic acid.
a)What is the [H+] concentration of this buffer?
Ans: pH=5 = -log[H+]
Concentration of [H+] = 1.00 *10^-5

b)What is the ratio [acetic acid]/[acetate]?
Ans: Ratio of [acetic acid] / [acetate] = 0.555

c)How many moles of sodium acetate (NaC2H3O2) must be added to 1 l of acetic acid 0.1 M to prepare this buffer?
0.1802mole of NaC2H3O2 used to added to 1litre of 0.1M acetic acid to prepare a pH5 buffer solution.