Chemistry question?
The solubility product constant (Ksp) for the dissolution of Mg(OH)2 as represented by the chemical equation is 1.8x10^-11.
Mg(OH)2 (s) <==> Mg^2+ (aq) + 2OH^- (aq)
Calculate the grams of Mg(OH)2 that dissolves in 1300. mL of water.
Thanks.
Answer:
. The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 x 10¯17 at represented by the equation above. Write the expression ...
Let x = moles/l Mg(OH)2 that dissolve . This will produce x mol/L Mg2+ and 2x moles/L OH-
The equilibrium requires that
Ksp = [ Mg2+] [OH-]^2 = x (2x)^2 = 4 x^3
1.8 x 10^-11 = 4x^3
x = 0.000165 M This gives how many moles Mg(OH)2 can dissolve per liter.
To get grams per 1300 mL = 1.300 L we need to multiply by the weight of 1 mole and 1.300
0.000165 mol/L x 58.3 g/mol x 1.3 L = 0.012 g
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