1. A solution of magnesium hydroxide is prepared by dissolving 361.824 mg of Mg(OH)2 in 273.5...

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1. A solution of magnesium hydroxide is prepared by dissolving361.824 mg of Mg(OH)2 in 273.5 mL of water. What is the equilibriumconcentration of OH- in this solution?
For magnesium hydroxide, Ksp = 7.1 x 10-12.
Assume no volume change.
2. The above solution is filtered to remove all undissolvedmagnesium hydroxide solid. What mass of solid Mg(OH)2 should berecovered from the solution?
Express your answer in units of mg.
3. After the above solution was filtered and all undissolvedmagnesium hydroxide solid removed, 564.1 mg of tin chloride isadded.
SnCl2 completely dissociates in aqueous solution. Hydroxide reactswith tin (II) to form a Sn(OH)2complex.
What is the equilibrium concentration of Sn2+ in thissolution?
The formation constants for the tin-hydroxide complexes are:Kf1 = 2.5 x 1010, Kf2 = 3.2 x 1013.
4. What is the equilibrium concentration of the SnOH+ complex inthis same solution?
5. What is the equilibrium concentration of Sn(OH)2 in this samesolution?
6. What is the ionic strength of the above solution?

Answer & Explanation Solved by verified expert

4.2 Ratings (497 Votes)

1.
Mg (OH)2 <-----> Mg2+ + 2OH-
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â sÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 2s
Ksp = [Mg2+] [OH-]^2
7.1*10^-12 = s * (2s)^2
7.1*10^-12 = 4*s^3
s = 1.21*10^-4 M
[OH-] = 2s
Â Â Â Â Â Â Â Â Â Â Â Â Â = 2* 1.21*10^-4
Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.42*10^-4 M
Answer: 2.42*10^-4 M

2.
Mg(OH)2 dissolved = 1.21*10^-4 mL
V = 273.5 mL = 0.2735 L
number of moles of Mg(OH)2 = M*V
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.21*10^-4 * 0.2735
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 3.31*10^-5 mol
So,
amount of Mg(OH)2 dissolved = number of moles * molar mass
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (3.31*10^-5) * 58.32
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.931*10^-3 g
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1.931 mg
Amount of Mg(OH)2 to be recovered = 361.824 mg - 1.931 mg = 359.893 mg
Answer: 359.893 mg

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