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

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!!!

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.

OHHH like a limiting reagent problem!

thankk you!!!!

When calculating the pH of a solution resulting from the addition of an acid and a base, you need to use the concept of stoichiometry, which is based on the balanced chemical equation.

In this case, the balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl -> H2O + NaCl

Since NaOH is a strong base and HCl is a strong acid, they dissociate completely in water. Therefore, the net ionic equation is:
OH- + H+ -> H2O

We are given that the initial concentration of NaOH is 0.10 M and the volume added is 10 mL, which can be converted to 0.010 L. We can calculate the initial moles of NaOH as:
0.10 M x 0.010 L = 0.001 mol NaOH

Similarly, the initial concentration of HCl is 0.10 M and the volume is 90 mL, which can be converted to 0.090 L. The initial moles of HCl can be calculated as:
0.10 M x 0.090 L = 0.009 mol HCl

Now, focusing on the step you mentioned, the reason why we subtract the moles of NaOH from the moles of HCl is because NaOH reacts with HCl in a 1:1 stoichiometric ratio. This means that for every 1 mole of NaOH, 1 mole of HCl is consumed. By subtracting the moles of NaOH from the moles of HCl, we determine the net moles of HCl remaining after the reaction.

So, in this case, the net moles of HCl remaining after the reaction is:
0.009 mol HCl - 0.001 mol NaOH = 0.008 mol HCl

Next, we need to calculate the new molarity of the HCl solution. This can be done by dividing the net moles of HCl by the new total volume of the solution, which is 0.10 L (since the volumes of NaOH and HCl are added together):
(0.008 mol HCl) / 0.10 L = 0.08 M HCl

Now that you have the new molarity of the HCl solution, you can proceed to calculate the pH using the appropriate equations or calculator.

Remember, when dealing with acid-base reactions, it is essential to consider the stoichiometry of the reaction and the resultant concentrations of the reactants involved.