Question

300ml 0.4M Pb(NO3)2 solution is mixed with 50 ml 0.6 M Ba3(PO4)2. The Pb2+ concentration after the precipitation of Pb3(PO4)2 is​

Answer 1

chemical equation Pb3(PO4)2 + 3Ba(NO3)2 → Ba3(PO4)2 + 3Pb(NO3)2

v 1 = 300 ml        M 1 = 0.4M

V 2 = 50 ml        M2 = 0.6 M

M result  = M1V1 + M2V2/V1 +V2

              0.4*300 + 0.6*50/350  = 120+30/350 = 150/350 =0.4 M

    Pb3(PO4)2 + 3Ba(NO3)2 → Ba3(PO4)2 + 3Pb(NO3)2

    1 mole                 3 mole      1 mole                 3 mole

       0.4M                  0.6M                                     0.4 M

Pb3(PO4)2 → Pb3 + (PO4)2

 0.4 M          ?

M1 = 0.4    V 1 = 300ML

M2 = ?       V 2  =  350          

M1V1=M2V2

0.4 * 300/350 =M2

M2 =0.34 M

THERE FORE [ pb2+ ] = 0.34 M