What is the pH of the solution created by combining 12.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?

with 8.00 mL of the 0.10 M HC2H3O2(aq)?

mL NaOH pH wHCl pH wHC2H3O2
12.30

What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?

mL NaOH pH wHCl pH wHC2H3O2
12.30

2 answers

NaOH + HCl ==>NaCl + HOH
Strong base + strong acid.
calculate mols NaOH. M x L = ??
calculate mols HCl. M x L = ??
Determine mols NaCl and HOH produced.
Determine excess NaOH OR HCl and pH from that.

For NaOH + HC2H3O2 ==> NaC2H3O2 + HOH
Strong base + weak acid.
Determine mols NaOH. M x L = ??
Determine mols HC2H3O2. M x L = ??
Determine mols NaC2H3O2 formed.
Determine mols HC2H3O2 in excess.
This gives you a weak acid and its salt which is a buffer solution. Determine pH from Henderson-Hasselbalch equation which is pH = pKa + log (base)/(acid).

Post your work if you get stuck.
mols NaOH = 0.1 mol/L x 0.0123 L = 0.00123 mols

mols HCL = 0.1 mol/L x 0.008 L = 0.0008

how do you determine mols NaCl and HOH produced?