[usenet69]

Below I wil post my titration Question, followed by my effort to solve it. I got the wrong answer, and do not know where I made my mistke.

A 210 mL sample of rainwater is titrated using .0095M NaOH solution. If 4.7 mL of the NaoH solution is required to reach teh endpoint,  what was the pH of the water?

So fist I set it up like a equation:

N x 210 = .0095 x 47

for n, I got .02126

-log of 02126 is 2.67239


I don't see where I went wrong?

[Mitch]

It's not an M1V1 = M2V2 question.

You first need to calculate the moles of NaOH used, which will equal the moles of acid present, and I'll assume you can take it from there.

[Demotivator]

The setup is right but the calc is wrong.
.0095 x 4.7 = .04465
n = .0002126

[AWK]

This is not excelent problem. rain water contains mixture of strong and weak acids. To calculate pH of rain water the assumption should be make that only strong acid is present.