1) In a titration of a 100.0mL 1.00M HCl strong acid solution with 1.00M NaOH, w
ID: 1017905 • Letter: 1
Question
1) In a titration of a 100.0mL 1.00M HCl strong acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 33.1 mL of NaOH? (3 significant figures)
**Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4**
2) In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 38.6 mL of NaOH? Ka = 1.80 x 10-5 for HA. (3 significant figures)
**Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4**
3) Which of the following occurs when HCl is added to a buffer containing (CH3)3N and (CH3)3NH+?
Select all that apply.
- The concentration of (CH3)3N will increase
- The concentration of (CH3)3NH+ will increase
- The concentration of (CH3)3N will decrease
- The concentration of (CH3)3NH+ will decrease
Explanation / Answer
1)
moles of HCl present = M*V = 1.00 M * 100.0 mL = 100.0 mmol
moles of NaOH added = M*V = 1.00 M * 33.1 mL = 33.1 mmol
33.1 mmol of each will react to form salt and water
After reaction, moles of HCl remaining = 100.0 mmol - 33.1 mmol = 66.9 mmol
Total volume = 100.0 mL + 33.1 mL = 133.1 mL
Final concentration of HCl,
[HCl] = number of moles / total volume
= 66.9 / 133.1
=0.5026 M
pH = -log [H+]
= -log (0.5026)
= 0.299
Answer: 0.299
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