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feven feven
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11 years ago
Given that 20.00 mL of phosphoric acid solution required 15.50 mL of 0.200M
NaOH for the first equivalent point. What is the molarity of the phosphoric acid?
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#1
11 years ago
Use the formula:-
M1V1=M2V2
M1= molarity of 1st solution
V1= volume of 1st solution
M2= molarity of 2nd solution
V2= volume of 2nd solution
Here,
M1= ?
V1= 20mL
M2= .2 M
V2= 15.50mL
Putting in values we get,
M1*20 = .2*15.5
M1= .2*15.5/20 = .155M
So the molarity of phosphoric acid is 0.155M
wrote...
#2
11 years ago
NaOH + H3PO4 = NaH2PO4 + H20
The first equivalent point
Mol(NaOH) = 0.2 x 15.5  / 1000 =>
Mol(NaOH) = 0.0031 mol
Since this number of moles NaOH gives NaH2PO4
Then moles of NaH2PO4 used is 0.0031
Molarity (H3PO4) = 0.0031 x 1000 / 20.0 = 0.155M
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