Asolution is prepared by mixing 150.0 ml of 1.00*10^-2 m mg(no3)2 and 250.0 ml of 1.00*10^-1m naf calculate the concentrations of mg2+ and f at equilibrium with solid mgf2 ksp= 6.4*10^-9

First, we have to get moles of Mg2+ = molarity * volume 

                                                              = 1 x 10^-2 * 0.15 L

                                                              = 0.0015 mol

initial [Mg2+] = 0.0015 / 0.4 L(total volume)

                      = 0.00375 M

then, moles of F- = molarity * volume 

                             = 1 x 10^-1 * 0.25

                             = 0.025 Mol

initial [F-] = 0.025 / 0.4 = 0.0625 M

by using ICE table:

                  MgF2(s)  →         Mg2     +             2F-        

initial                                  0.00375                 0.0625

change                                    + X                          +2 X

Equ                                    0.00375+X           0.0625+2X
        

when Ksp = [Mg2+][F-]^2 

by substitution: 

6.4 x 10^-9 = (0.00375+X)( 0.0625 +2X)^2   by solving for X

∴X=  0.00024  M 

∴[Mg2+] = 0.00375 + X

               = 0.00375 + 0.00024

              =  0.004 M 

and [F-] = 0.0625 + 2X

             = 0.0625 + 2* 0.00024

             = 0.06 M

Hno3(aq) → h+(aq) + no3−(aq). picture ii best represents the strong acid hno3. strong acids are strong electrolytes

Hey how did you manage to type such a big lab report