# Problem: 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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###### FREE Expert Solution

A. First, calculate the amount of acid and base that will react

molarity (volume) → moles

Recall:

*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

Reaction:

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 ###### Problem Details

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)