If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the phosphoric a

Question

If 15.0 mL of phosphoric acid completely neutralizes 38.5 mL of 0.150 mol/L calcium hydroxide, what is the concentration of the phosphoric acid?

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Hưng Khoa 18 mins 2021-07-22T16:47:54+00:00 1 Answers 0 views 0

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    2021-07-22T16:48:56+00:00

    Answer:

    Let me give it a try.

    H3PO4 + Ca(OH)2 = Ca3(PO4)2 + H2O

    Balancing this reaction

    2H3PO4 + 3Ca(OH)2 == Ca3(PO4)2 + 6H2O.

    Moles= Molarity x Volume

    Volume = 38.5ml = 0.0385L

    Moles of Ca hydroxide = 0.150m/L x 0.0385L

    (Notice the units canceling out…leaving moles).

    =0.005775moles of Ca(OH)2.

    From balanced reaction…

    3moles of Ca(OH)2 completely reacts with 2moles of H3PO4

    0.005775moles of Ca(OH)2 would completely react with….

    = 0.005775 x 2/(3)

    =0.00385moles of H3PO4.

    Now we’re looking for its Concentration in Mol/L

    Molarity=Moles of solute/Volume of solution(in L)

    Volume of solution assuming no other additions to the reaction = 15ml + 38.5ml =53.5ml =0.0535L

    Molarity = 0.00385/0.0535

    =0.072Mol/L.

    If this is wrong

    then Simply Try The formula for Mixing of solutions

    C1V1 = C2V2

    0.15 x 38.5 = C2 x (15+38.5)

    C2 = 0.11M/L.

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