ap chemistry
posted by jenna .
what volume of 0.0100 M NaOH must be added to 1.00 L of 0.0500 M HOCl to acheive a PH of 8.00?

1000 mL x 0.05M = 50 millimols.
x mL x 0.1M = ? mmols.

............HOCl + OH^ ==> OCl^ + H2O
initial.....50......0........0........0
add.................x................
change......x....x.........x........x
equil.......50x....0.........x........x
I used 3E8 for Ka for HOCl
pH = pKa + log(base)/(acid)
8.00 = 7.52 + log(x)/(50x)
Solve for x = millimols OH^ added. Then mL = mmols/M. I get approximately 380 mL 
I believe you made a slight error in the last calculation from mmol > mol. The answer should be around 3.8L.

Ka=3.5X10^8
pH+pOH=14
148.00=pOH
pOH=log[OH], so [OH]=10^(6.00)=1x10^6
HOCl + OHH20 + OCl
.05M X 0
X X +X
.05X 0 X
THEN USE HENDERSONHASSELBACK EQUATION
pH=pKa+LOG[A/HA]
8.00=LOG(3.5X^8)+ LOG [X/ (.05X)]
8.00=7.46+LOG [X/(.05X)]
.544=LOG[X/(.05X]
TO GET RID OF LOG, RAISE 10^.544
3.5=X/(.05X)
3.5(.05X)=X
.1753.5X=X
.175=3.5X+X
.175=X(3.5+1)
.175=4.5X
X=.0388 mol
.0388mol x (1 L/ .100 mol)=3.9 L