[shotgunrebuttal]

I was given the following problem:

Suppose 100.0 mL of 2.00 x 10^-7 M hydrobromic acid is mixed with 170.0 mL of 1.20 x 10^-7 M LiOH.  What is the pH?

I did the problem out and all I could come up with was a pH of 5.17, which doesn't really make sense to me since it is the LiOH that is in excess.

[AWK]

Show your work. Your result is, in fact, wrong. Calculate an exces of base, OH- concentration, then pOH and finally pH.

[Borek]

You have to account for water autodissociation.

[AWK]

Borek is, of course, right.
You should use an eqution of type x(x+c)=Kw where x is the OH- or H+ concnetration from water and c is the excess OH- concnetration after neutralization.

[shotgunrebuttal]

Okay so:

2.00 x 10^-8 mols HBr
minus
2.04 x 10^-8 mols LiOH

is 4.00 x 10^-10 OH^-
/
.27 L
=
1.481 x 10^-9

1.00 x 10^-14 = [x ] [x + 1.481 x 10^-9
0 = x² + 1.481 x 10^-9x - 1.00 x 10^-14
x = 9.93^-8

-log(9.93^-8) = pH = 7.003

Does that look right?

[Borek]

Does that look right?

Yes and no.

Yes, as its a correct method and pH 7.00 is a correct answer.

No, as 9.93^-8 is not 9.93x10^-8