Hi Mam,
I was wondering how you would approach this problem?Screenshot 2026-04-17 180144.png
HCl + NaOH → NaCl+ H2O
Step 1: Calculate mmoles of each reactant
mmol of HCl=M×mL=0.25×10.0=2.50 mmol
mmol of NaOH=M×mL=1.5×5.00=7.50 mmol
Step 2: Identify the limiting and excess reactant
Since the reaction is 1 : 1:
HCl = 2.50 mmol
NaOH = 7.50 mmol
So, HCl is the limiting reactant and NaOH is the excess reactant.
Step 3: Use an initial–final table in mmoles
HCl + NaOH → NaCl+ H2O
Initial 2.50 7.50 0 0
Final 0 5.00 2.50 2.50
After the reaction, 5.00 mmol of NaOH is left in excess.
That means the solution is basic, so calculate [OH⁻].
Step 4: Calculate the concentration of excess OH⁻
Total volume after mixing: 10.0 mL+5.00 mL=15.0 mL
[OH −]= 15.0 mL /5.00 mmol
=0.333 M
Step 5: Find pOH and pH
pOH=−log(0.333)=0.48
pH=14.00−0.48=13.52