# ap chemistry

posted by on .

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?

• ap chemistry - ,

1000 mL x 0.05M = 50 millimols.
x mL x 0.1M = ? mmols.
-----------------
............HOCl + OH^- ==> OCl^- + H2O
initial.....50......0........0........0
change......-x....-x.........x........x
equil.......50-x....0.........x........x
I used 3E-8 for Ka for HOCl
pH = pKa + log(base)/(acid)
8.00 = 7.52 + log(x)/(50-x)
Solve for x = millimols OH^- added. Then mL = mmols/M. I get approximately 380 mL

• ap chemistry - ,

I believe you made a slight error in the last calculation from mmol --> mol. The answer should be around 3.8L.

• ap chemistry - ,

Ka=3.5X10^-8
pH+pOH=14
14-8.00=pOH
pOH=-log[OH-], so [OH-]=10^(-6.00)=1x10^-6
HOCl + OH--------H20 + OCl-
.05M X 0
-X -X +X
.05-X 0 X
THEN USE HENDERSON-HASSELBACK EQUATION
pH=pKa+LOG[A-/HA]
8.00=-LOG(3.5X^-8)+ LOG [X/ (.05-X)]
8.00=7.46+LOG [X/(.05-X)]
.544=LOG[X/(.05-X]
TO GET RID OF LOG, RAISE 10^.544
3.5=X/(.05-X)
3.5(.05-X)=X
.175-3.5X=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