in an acid base titration, a sample of unknown concentration phosphoric acid is

Question

in an acid base titration, a sample of unknown concentration phosphoric acid is analyzed by titration with a solution of sodium hydroxide solution. In this analysis just enough sodium hydroxide solution of known molar concentration is added to just react with all of the phosphoric acid. When this condition has been met, the endpoint of the titration has been reached. suppose that 26.38 mL of a 0.100 M sodium hydroxide solution is added to a 30.00 mL sample of the phosphoric acid solution results in complete neutralization. H3PO4(aq) +3NaOH(aq) = 3H2O(l) + Na3PO4(aq)

a) how many moles of sodium hydroxide is added to the phosphoric acid solution after the titration

b) determine the molarity of the unknown phosphoric acid solution

c) in this reaction the sodium ion is a spectator ion and remains in the solution. how many moles of sodium ion are there in solution after the titration is completed? what would be the molarity of sodium ion?

Explanation / Answer

a) we know that

moles = molarity x volume (L)

so

moles of NaOH = 0.1 x 26.38 x 10-3

moles of NaOH added = 2.638 x 10-3

b)

now

consider the reaction

H3P04 + 3 NaOH --> 3 H20 + na3P04

we can see that

moles of NaOH added = 3 x moles of H3P04

so

2.638 x 10-3 = 3 x moles of H3P04

moles of H3P04 = 0.879333 x 10-3

now

molarity = moles / volume (L)

so

molarity of H3P04 = 0.879333 x 10-3 / 30 x 10-3

molarity of H3P04 = 0.0293

c)

all the sodium ions comes from NaOH

NaOH --> Na+ + OH-

so

moles of Na+ = moles of NaOH added

so

moles of Na+ = 2.638 x 10-3

now

final volume = 30 + 26.38

final volume = 56.38

now

molarity = moles / volume

so

molarity = 2.638 x 10-3 / 56.38 x 10-3

molarity of Na+ = 0.0468

so

the molarity of Na+ ions is 0.0468

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