Calculate the pH from the addition of 10 mL of a 0.10 M NaOH solution to 90 mL of 0.10 M HCl.

Question

okay so i understand how to do most of it but i get messed up at one part..

so i made my equation:
NaOH + HCl -> H2O + NaCl

(strong base, strong acid= complete dissociation)

so net is: OH + H -> H2O

i found the mol of NaOH which is 0.001
mol of HCl is 0.009

the next step is where i mess up...in the problem it says "addidtion of 10 mL..." in the actual solution you subtract the moles over the liters to get the new molarity..like this:
(0.009 mol HCl - 0.001 mol NaOH)/0.10 L

and that gives me 0.08..
i know how to calculate the ph from here ..but in the step i mentioned why is it subtracting moles intsead of adding them??

Thankk youu!!!

Answer 1

This may help.
............NaOH + HCl ==> NaCl + H2O
initial....0.001..0.009.....0......0
change...-0.001..-0.001....+0.001..0.001
equil.......0.....0.008.....0.001.0.001
The base and acid react. So 0.001 moles NaOH "disappear" and take 0.001 moles HCl with it (to make 0.001 mole H2O and NaCl).
Now you have 0.008 moles HCl (no NaOH) and M = moles/L = 0.008/0.10 = 0.08M and from there pH.

Answer 2

OHHH like a limiting reagent problem!
thankk you!!!!