A triprotic acid requires three moles of base to neutralize it, and the protons
ID: 821734 • Letter: A
Question
A triprotic acid requires three moles of base to neutralize it, and the protons are removed one at a time as follows:
H3A + OH- <-> H2A- + H2O
H2A- + OH- <-> HA 2- + H2O
HA 2- + OH- <-> A 3- + H2O
You have a 40.0 mL solution of a triprotic acid, H3A, with a concentration of 0.0588 M. You titrate it with a 0.250 M solution of NaOH. How many millimoles of H3A are you titrating? (Give units)
How many millimoles of NaOH are required to complete the titration? (Give units)
What volume of NaOH will be needed to completely titrate the acid?
What volume of NaOH will be needed to reach the first equivalence?
What volume of NaOH will be needed to reach the second equivalence point?
Explanation / Answer
H3A + OH- <-> H2A- + H2O
H2A- + OH- <-> HA2- + H2O
HA2- + OH- <-> A3- + H2O
(1) Millimoles of H3A = volume x concentration of H3A
= 40.0 x 0.0588 = 2.352 mmol = 2.35 mmol
(2) Millimoles of NaOH (complete titration) = 3 x millimoles of H3A
= 3 x 2.352 = 7.056 mmol = 7.06 mmol
(3) Volume of NaOH (complete titration) = millimoles/concentration of NaOH
= 7.056/0.250 = 28.22 mL = 28.2 mL
(4) Volume of NaOH (first equivalance) = 1/3 x volume of NaOH (complete titration)
= 1/3 x 28.22 = 9.41 mL = 9.4 mL
(5) Volume of NaOH (second equivalance) = 2/3 x volume of NaOH (complete titration)
= 2/3 x 28.22 = 18.82 mL = 18.8 mL
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