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Question

Old bottle of phosphoric acid that is labelled “0.6829M”.

a. A standardized bottle of sodium hydroxide with a known concentration of 0.8573M. How many milliliters of this NaOH solution should be required to reach first equivalence point use to titrate 26.00mL of the old phosphoric acid? How much would be required to reach the second equivalence point? The third equivalence point?

b. When you perform the titration, you reach the second equivalence point when 37.81mL of 0.8573M NaOH(aq) is added. What is the actual concentration of the old phosphoric acid based upon this titration?

Answer 1

The equivalence points should be when the OH^- from the NaOH neutralizes the H⁺ from the acid. So the first point will require a volume of NaOH, the 2nd, twice as much, and the third, three times as much. So you can use the the titration equation for a monoprotic acid and NaOH:

M_AV_A = M_BV_B (which is equating n_H with n_OH

V_B = M_AV_A/M_B = .6829 M*26ml/.8573M = ml of standard NaOH

For the second part you can use the V of base required, the V of base that should be used for the quoted molarity, and the quoted molarity in order to find the actual concentration of acid:

M_A' = M_A(V_B')/V_B where prime is what you actually have and unprimed is the calculated assuming the given M_A (Don't forget to use the V_B for the second point.

Answer 2

H3PO4 + NaOH ->

see video