200.0 ml of 0.5000 M acetic acid is added to 200.0 ml of 0.5000 M NaOH. What is

Question

the final pH? The Ka of acetic acid at 25 oC is 1.76 x 10-5.

n(NH3COOH)=CxV=0.5x200/1000=0.1mol
n(Naoh)=CxV=0.5x200/1000=0.1mol

I have no idear of what to do after finding the number of moles

Answer 1

use the h-h equation : pH=pKa+log(base/acid)
since you're given ka,calculate kb using the formula:
kw/ka=kb
1*10^-14/1.76*10-5=5.68*10^-10.
pKb will therefore be -log(5.68*10^-10)=9.25
pH=9.25+log(0.5/0.5)=9.25