PART A: mass KHP (g): 0.7621 g initial buret reading: 0.00 mL final buret readin

Question

PART A:
mass KHP (g): 0.7621 g
initial buret reading: 0.00 mL
final buret reading: 14.95 mL
vol of NaOH used: 14.95 mL
Molarity NaOH: 0.2496 M

mass KHP (g): 0.7549 g
initial buret reading: 14.95 mL
final buret reading: 29.70
vol of NaOH used: 14.75 mL
Molarity NaOH: 0.2506 M

The standardized base from Part A is used to determine the molarity of a polyprotic acid. A 15.00 mL sample of acid is titrated with the standard base. If the buret readings are Vi=0.25mL and Vf= 30.15 mL, what is the concentration of the unknown acid if its structure is:

1) HA

[HA] =_________ M

2) H2A

[H2A] = __________ M

3) H3A

[H3A] = __________ M

Please show steps to solution if you can! Thank you!

Explanation / Answer

Average molarity of NaOH = (0.2496 + 0.2506)/2 = 0.2501 M

Volume of NaOH = Vf - Vi = 30.15 - 0.25 = 29.90 mL

Moles of NaOH = volume x molarity of NaOH = 29.90/1000 x 0.2501 = 0.00747799 mol

(1) HA + NaOH => NaA + H2O

Moles of HA = moles of NaOH = 0.00747799 mol

[HA] = moles/volume of HA = 0.00747799/0.01500 = 0.4985 M

(2) H2A + 2 NaOH => Na2A + 2 H2O

Moles of H2A = 1/2 x moles of NaOH = 1/2 x 0.00747799 = 0.003738995 mol

[H2A] = moles/volume of H2A = 0.003738995/0.01500 = 0.2493 M

(3) H3A + 3 NaOH => Na3A + 3 H2O

Moles of H3A = 1/3 x moles of NaOH = 1/3 x 0.00747799 = 0.002492663 mol

[H3A] = moles/volume of H3A = 0.002492663/0.01500 = 0.1662 M

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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