If 25mL of 2.00M NAOH is added tpo 50mL of 0.10M HC2H3O2 (Ka=1.8*10^-5), what is the pH?

Question

moles HC2H3O2= c*v = .1 * .05L = .005

moles NaOH= c*v = .2 * .025L = .005

pH= Pka + log (B/A)
= 4.75 + log (.005/.005)
= 4.75

I'm not sure how to calculate the moles I know i'm making a mistake there

Answer 1

Moles=concentration*volume

You got the moles wrong on NaOH

Answer 2

n = c* v

= .2mol/L * .025L = .005

is the c im using wrong or the V?

Answer 3

The original post says 25 mL of 2.00 M NaOH = 25 x 2 = 50 millimoles or 0.05 moles NaOH. But your solution says 0.2 x 0.025 = ??. Something is awry. Is it 2.00 M or 0.200 M? I think that's the trouble spot.

Answer 4

oh no that was a type it's actually .20M NaOH

so that gives me: n = c* v

= .2mol/L * .025L = .005

moles HC2H3O2= c*v = .1 * .05L = .005

I still get the same moles for both of them!

Answer 5

So, what do you have in the solution?
You have added 0.005 moles NaOH to 0.005 moles HC2H3O2. That produces 0.005 moles NaC2H3O2 (sodium acetate, the salt which is now dissolved in 75 mL water) + H2O from the neutralization reaction with no acid or base remaining unreacted. Your use of the H-H equation isn't justified BECAUSE you don't have a buffer solution. All you have is the salt of a weak acid and a strong base, namely NaC2H3O2.
The concn of the salt is 0.005 moles/0.075 mL. So it will hydrolyze as
(I like to call acetate Ac^----it sure saves a lot of typing).
Ac^- + HOH ==> HAc + OH^-
Kb = (Kw/Ka) = (OH^-)(HAc)/(Ac^-)
You know Kw, Ka, and (Ac^-). Solve for (OH^-) and pOH and pH from that.