8. Titration 28.85 mL of 0.1000 M NaOH is required to titrate 0.129 g of an unkn

Question

8. Titration 28.85 mL of 0.1000 M NaOH is required to titrate 0.129 g of an unknown monoprotic acid. What is the molecular weight of the unknown acid? 28.85 mL of 0.1000 M NaOH is required to titrate 0.129 g of an unknown diprotic acid What is the molecular weight of the unknown acid? Hint: Look at page 259 for guidance. What do you notice about the molecular weights of the two unknown acids? Note that the same volume of the same concentration of base was used to titrate the same mass of two different unknown acids.

Explanation / Answer

HA + NaOH ---------> NaA + H2O
no of moles of NaOH = molarity * volume in L
                      = 0.1*0.02885
                      = 0.002885 moles
from the balanced equation
1 mole of NaOH react with 1 mole of HA
0.002885 moles of NaOH react with 0.002885 moles of HA
molecular weight of HA = weight of HA/no of moles of HA
                        = 0.129/0.002885   = 44.7g/mole

H2A + NaOH ---------> NaHA + H2O
no of moles of NaOH = molarity * volume in L
                      = 0.1*0.02885
                      = 0.002885 moles
from the balanced equation
1 mole of NaOH react with 1 mole of H2A
0.002885 moles of NaOH react with 0.002885 moles of H2A
molecular weight of H2A = weight of H2A/no of moles of H2A
                        = 0.129/0.002885   = 44.7g/mole

H2A + 2NaOH ---------> Na2A + 2H2O
no of moles of NaOH = molarity * volume in L
                      = 0.1*0.02885
                      = 0.002885 moles
from the balanced equation
2 mole of NaOH react with 1 mole of H2A
0.002885 moles of NaOH react with 0.002885*1/2    = 0.0014425moles of H2A
molecular weight of H2A = weight of H2A/no of moles of H2A
                         = 0.129/0.0014425   = 89.4g/mole >>>>answer

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.