chem refresher?

Question

49 mL of .200 M HCl is mixed with 50 mL of .2 M NaOH to reach endpoint.

Calculate the [H+] and [OH-]

Answer 1

49mL HCl x 0.200molHCl/1000mL HCl = 9.8 x 10^-3 mol HCl

50mL NaOH x 0.200molNaOH/1000mL NaOH = 10 x 10^-3 molNaOH

So besides the NaCl formed, there would be 0.2 mole NaOH in 99mL solution.

0.2molNaOH/99mLsoln = 2 x 10^1-3 M OH- (Answer to part 1)

Kw = [H+][OH-] = 10^-14

(2 x 10^-3)[H+] = 10^-14

[H+] = 5 x 10^-10 M (Answer to part 2)