🤓 Based on our data, we think this question is relevant for Professor Delgado's class at FIU.
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)
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Strong Acid Strong Base Titrations concept. You can view video lessons to learn Strong Acid Strong Base Titrations. Or if you need more Strong Acid Strong Base Titrations practice, you can also practice Strong Acid Strong Base Titrations practice problems.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Delgado's class at FIU.