Use the information provided for each compound to answer the questions that foll

Question

Use the information provided for each compound to answer the questions that follow.

Silver bromate, Ksp=5.5x10-5

Silver sulfide, Ksp=6.0x10-51

Magnesium carbonate, Ksp=3.5x10-8

Manganese hydroxide, Ksp=1.6x10-13

a. Calculate the solubility (S) of each compound

b. Calculate the solubility product constant (Ksp) in terms of pKsp for each compound (use the correct number of significant figures). Show the formula you will use for the calculation.

c. At what pH would manganese hydroxide precipitate in a solution in which [Mn(OH)2] = 2.3x10-4 M? (Show all your work, use the correct number of significant figures)

Explanation / Answer

a i.

AgBrO3 <-> Ag+ + BrO3-

Ksp = [Ag+][BrO3-]

5.5*10^-5 = S*S

S = sqrt(5.5*10^-5) = 0.007416 mol of AgBrO3 per liter

a ii.

Ag2S<-> 2Ag+ + S-2

Ksp = [Ag+]^2[S-2]

5.5*10^-5 = (2S)^2*S

4*S^3 = 5.5*10^-5

S = ((5.5*10^-5)/4 ) ^(1/3) = 0.023957 ol of Ag2S per liter

MgCO3 <-> Mg+2+ CO3-2

S = sqrt(Ksp) = sqrt(3.5*10^-8) = 0.0001870 mol of MgCO3 per liter

Mn(OH)2 <-> Mg+2 + 2OH-

Ksp = [Mn+2][OH-]^2

1.6*10^-13= S* (2S)^2

4*S^3 = 1.6*10^-13

S = ((1.6*10^-13)/4)^(1/3) = 0.0000341995 mol of Mn(OH)2

b)

for

AgBrO3 = -log(Ksp) = -log(5.5*10^-5) = 4.259637 --> 2 sig fig 4.3

for

Ag2S = -log(Ksp) = -log(6*10^-51) = 50.22 --> 2 sig fig 50

for

MgCO3= -log(Ksp) = -log(3.5*10^-8) = 7.45 --> 2 sig fig 7.5

for

Mn(OH)2= -log(1.6*10^-13) = 12.7958--> 2 sig fig 13

c)

find pH in which

[Mn(OH)2] = 2.3*10^-4

[Mn+2] = 2.3*10^-4

[OH-] = ?

1.6*10^-13 = [Mn+2][OH-]^2

[OH-] = sqrt((1.6*10^-13)/(2.3*10^-4)) = 0.00002637521

pOH 0 -log(0.00002637521) = 4.578

pH = 14-poH = 14-4.578 = 9.422

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