How many grams of a solid magnesium chloride could you add to 1L of 0.25M NaOH s

Question

How many grams of a solid magnesium chloride could you add to 1L of 0.25M NaOH solution before a permanent precipitate formed? (Ksp of magnesium hydroxide is 1.8X10-11 @ 25C)

Explanation / Answer

MgCl2 + 2 NaOH -----> Mg(OH)2 + 2 NaCl . . . or Mg(2+) + 2 HO(-) -----> Mg(OH)2 M(MgCl2) = M(Mg) + 2M(Cl) M(MgCl2) = 24,3 + (2 x 35,45) M(MgCl2) = 95,2 g/mole Ksp = [ Mg(2+) ] x [ HO(-) ] ² = 1,8•10^-11 [ Mg(2+) ] = Ksp / [ HO(-) ] ² [ Mg(2+) ] = 1,8•10^-11 / 0,25 ² [ Mg(2+) ] = 2,88•10^-10 mole/L [ Mg(2+) ] = n(Mg(2+)) / V [ Mg(2+) ] = m(Mg(2+)) / ( M(MgCl2) x V) . . . finally : m(Mg(2+)) = [ Mg(2+) ] x M(MgCl2) x V m(Mg(2+)) = 2,88•10^-10 x 95,2 x 1 m(Mg(2+)) = 2,74•10^-8 g of MgCl2 I hope to have answered your question.

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