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Title: chemistry help?,,, doing a lab on polyprotic acids and titration, please help? Post by: feven on May 9, 2013 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? Title: chemistry help?,,, doing a lab on polyprotic acids and titration, please help? Post by: tomtom1234 on May 9, 2013 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 Title: chemistry help?,,, doing a lab on polyprotic acids and titration, please help? Post by: Leendr on May 9, 2013 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 |