Will a precipitate of magnesium fluoride form when 200.0 ml of 1.9 x 10-3 M MgCl2...

Will a precipitate of magnesium fluoride form when 200.0 ml of 1.9 x 10^-3 M MgCl₂ are added to 300.0 ml of 1.4 x 10^-2 M NaF (K_sp (MgF₂ = 6.9 x 10^-9 )

Please explain, thanks!

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Expert Solution

Answer – We are given, 200.0 ml of 1.9 x 10^-3 M MgCl₂ and 300.0 ml of 1.4 x 10^-2 M NaF, , K_sp MgF₂ = 6.9 x 10^-9

We know when we added the MgCl₂ and NaF there is formed MgF₂, so there is formed precipitation or not we need to calculate the Q_sp

We know expression for the Q_sp

Q_sp = [Mg²⁺][F^-]²

Now we need to calculate the[Mg²⁺] and [F^-] after mixing the solution.

Total volume = 200 +300 = 500 mL

Moles of MgCl₂ = moles of Mg²⁺ = 1.9 x 10^-3 M * 0.200 L

= 0.00038 moles

Moles of NaF = moles of F^- = 1.4 x 10^-2 M * 0.300 L

= 0.0042 moles

So, [Mg²⁺] = 0.00038 moles / 0.500 L

= 0.00076 M

[F^-] = 0.0042 moles / 0.500 L

= 0.0084 M

So, Q_sp = [Mg²⁺][F^-]²

= (0.00076) (0.0084)²

= 5.36*10^-8

So, Q_sp > K_sp and we know when Q_sp > K_sp then there is precipitation occurred, so magnesium fluoride will a precipitate

queen_honey_blossom answered 1 year ago

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