Question

What is the pH of the solution created by combining 1.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?

This is how I would go about solving the problem, but could you tell me where I went wrong?

First I would find the number of mols of OH from NaOH and then use
HCl+OH -> HOH + Cl- to find the mols of HCl and Cl- after neutralization...and then use an ICE chart to find H and henceforth pH. I don't think this is correct because my answer came out negative.

Answer 1

The method looks ok.
NaOH + HCl ==> NaCl + HOH

mols NaOH = L x M = (1.30 mL/1000) x0.10 = 0.00013
mols HCl = L x M = (8/1000) x 0.10 = 0.0008

Reagent in excess is HCl. Therefore, all of the NaOH will be used, 0.00013 mols NaCl and HOH formed. Mols HCl remaining unreacted is 0.0008-0.00013 = 0.00067
Molarity HCl remaining is mols/L (combined volume is 1.13 + 8.9 = 9.13 mL)

Then find pH of the HCl solution. Should be about 1 or so.