A. First, calculate the amount of acid and base that will react
molarity (volume) → moles
*convert mL to L → 1 mL = 10-3 L
• 20.0 mL of 0.200 M HCl
• 30.0 mL of 0.200 M NaOH
Next, write the reaction equation between HCl and NaOH:
• HCl is a strong acid and based on the Bronsted-Lowry definition, an acid is a proton (H+) donor.
• NaOH is a strong base and based on Bronsted-Lowry definition, a base is a proton (H+) acceptor.
• HCl will donate a proton to NaOH
• The amount in moles of NaOH is greater than HCl. This means that HCl will be used up in the reaction while there will be some NaOH left unreacted
HCl(aq) + NaOH(aq) ⇌ NaCl + H2O(l)
(acid) (base) (salt) (water)
Determine how much NaOH will be left after the acid-base reaction
NaOH 6×10-3 mol
HCl 4×10-3 mol
2×10-3 mol NaOH
Determine the concentration of NaOH in the solution
Calculate the pH of the resulting solution if 20.0 mL of 0.200 M HCl(aq) is added to:
a. 30.0 mL of 0.200 M NaOH(aq)
b. 10.0 mL of 0.300 M NaOH(aq)
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