AP chemistry Unit 8 (Acids and Bases ) -Excess reactant and limiting reactant & pH

Chandrasekhar · April 17, 2026, 09:04:34 PM

Hi Mam,

I was wondering how you would approach this problem? Screenshot 2026-04-17 180144.png

uma · April 18, 2026, 09:37:56 AM

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

AP chemistry Unit 8 (Acids and Bases ) -Excess reactant and limiting reactant & pH

Chandrasekhar

uma