How many grams of acetic acid (a weak acid, CH3OOH with a Mw = 60 g/mole with a pKa = 4.75) are needed to chan?

Find initial H+ concentration.
pH=-log[H+]
11=-log[H+]
[H+]=1.0e-11

HC2H3O2 -> H+ + C2H3O2-
I___x____1.0e-11____0
C__-y______+y_____+y
E__x-y__1.0e-11+y__y

Find final [H+] concentration.
5.5=-log[H+]
[H+]=3.2e-6

Find 'y' in the table above.
1.0e-11+y = 3.2e-6
y=3.2e-6

HC2H3O2 -> H+ + C2H3O2-
I___x______1.0e-11____0
C_-3.2e-6__+3.2e-6___+3.2e-6
E_x-3.2e-6__3.2e-6___3.2e-6

Convert pKa to Ka.
4.75=-log(Ka)
Ka=1.78e-5

Set up equilibrium equation.
1.78e-5=(3.2e-6)^2/(x-3.2e-6)
x=3.78e-6 M

3.78e-6 = x/1000
x=0.00378 mol
0.00378 mol * (60 g/1 mol) = 0.227 g