Question

25.0mL of 0.1M Acetic acid is titrated to the equivalence point with 25.0mL of 0.1M NaOH. Calculate the pH at the equivalence point. Ka for acetic acid is 1.8 x 10^-5

Answer 1

I.) Since you have have equal concentration of both acetic acid and NaOH .025L, 0.1M for both, just use kb and [H], and ph

1.)ka= 1.8*10^-5

kb= 1.0*10^-14
1.8^-5
=5.67*10^-6

2.) [H]= 1.0*10^-14
sqrt5.67*10^-6*.025
=1.92*10^-9

3.) log(1.92*10^-9)
pH=8.72