Will a precipitate of magnesium fluoride form when 200. ml of 1.9 × 10–3 m mgcl2 are added to 300. ml of 1.4 × 10–2 m naf? [ksp (mgf2) = 6.9 × 10–9]

First, we have to get moles of MgCl2 = molarity * volume 

                                                              = 1.9 x 10^-3 M* 0.2L

                                                              = 3.8 x 10^-4 moles

and moles of naf = molarity * volume 

                             = 1.4 x 10^-2M * 0.3 L

                            = 4.2 x 10^-3 Moles

when the total volume = 0.2 L* 0.3L = 0.5 L

[Mg2+] = moles Mg2+ / total volume

            = 3.8 x 10^-4 mol / 0.5 L

            = 0.00076 M

[NaF] = moles NaF / total volume

         = 4.2 x 10^-3 mol / 0.5 L 

        = 0.0084 M

According to the reaction equation:

MgF2   ↔  Mg2+   +  2F-

when Qsp expression = [Mg2+] [F-]^2 

∴Qsp = (0.00076) * (0.0084)^2

          = 5.4 x 10^-8 

by comparing the value of Qsp and the Ksp value, we will find 

Qsp > Ksp 

∴ the precipitate of magnesium fluoride will form. 

                            

m=9

step-by-step explanation:

7m+8=71 subtract   8 from 71 which is 63 than divide it by 7m   equal 9.