1. ___ MgCl2(aq) + ____ NaOH(aq) -------> _____Mg(OH)2(s) + _____ NaCl(aq) a. Ba
ID: 991977 • Letter: 1
Question
1. ___ MgCl2(aq) + ____ NaOH(aq) -------> _____Mg(OH)2(s) + _____ NaCl(aq)
a. Balance the chemical equation above. MW magnesium hydroxide = 58.32 grams per mole.
b. Write the equilibrium solubility product(Ksp) in terms of magnesium and hydroxide ion concentration.
c. The solubility of magnesium hydroxide is 35.1 grams per mL. Calculate the Ksp.
2. Calcuim sulfate is slightly soluble in water (Ksp = 4.0 *10^-5)
a. Write the equilibrium solubility product(Ksp) in terms of calcium and sulfate ion concentration.
b. Calculate the solubility of calcium sulfate in terms of grams per mL. MW calcium sulfate = 136.14 grams per mole
Explanation / Answer
a.-) To balance the reaction you need to balance OH , writing 2 NaOH and then a number 2 in NaCl:
MgCl2(aq) + 2 NaOH(aq) ------->Mg(OH)2(s) + 2 NaCl(aq)
b.-) For Magnesium Hydroxide, you have the following equilibrium:
Mg(OH)2 Mg2+ + 2 OH- and Ksp = [Mg2+ ][OH-]2
c) Solubility of Mg(OH)2 is 35.1 g/mL, and we need this value in moles/L,so we have to use molar mass of magnesium hydroxide.(Notice: This solubility value is very high...Are you sure??? )
(35.1 g/mL )(1mol/58.32g)(1000ml/1L) = 601.9 moles/L
Then: Mg(OH)2 Mg2+ + 2 OH-
601.9M 1,203.8M
Then: Ksp = [Mg2+ ][OH-]2 = (601.9)(1,203.8)2 = 8.72x108 (Too big....)
2. Calcuim sulfate is slightly soluble in water (Ksp = 4.0 x10-5)
a. Write the equilibrium solubility product(Ksp) in terms of calcium and sulfate ion concentration.
CaSO4 Ca2+ + SO42- and Ksp = [Ca2+ ][SO42-]
If we consider "S" as solubility:
CaSO4 Ca2+ + SO42-
S S and Ksp = [Ca2+ ][SO42-] = (S)(S) 0 S2 .
Then 4.0 x 10-5 = S2
ans S = 6.3 x 10-3 moles/L
To convert moles/L to in g/L we use the molar mass:
S = (6.3 x 10-3 moles/L)(136.14g/1 mole) = 0.858g/L
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