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.
