Questions (Due July 9th by the END of class): For the following questions, you m

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Questions (Due July 9th by the END of class): For the following questions, you may need physical constants that aren't listed. They can be found either in your book or on the internet. Or, ask and we'll find them together. Mastery #1a: l put 10.00 grams of nickel (II) hydroxide in 500.0 mL of water and stir. After equilibrium has been established, tell me the following 1. 2. What (if any) mass of solid remains? What is the concentration (Molarity) of each chemical species in solution? Mastery #1b: l put 10.00 grams of nickel (ll) hydroxide and 10.00 g of barium hydroxide in sooo ml of water and stir. After equilibrium has been established, tell me the following: 1. What (if any) mass of solid(s) remains? 2. What is the concentration (Molarity) of each chemical species in solution? 3. Compare the nickel (ll) hydroxide results from the first and second problems. How are they the same? How are they different? Does the result make sense for problem 2 with respect to problem 1?

Explanation / Answer

#1a) Ksp or solubility product of Ni(II) hydroxide=1.6*10^-16

Ni(OH)2(s)<--->Ni2+ (aq) +2OH-(aq)

Ksp=[Ni2+][OH-]^2

Let [Ni2+]=S ,[OH-]=2S ,S=solubility of ions

Ksp=1.6*10^-16=S*(2S)^2=4S^3

S=3.42*10^-6 M

[Ni(OH)2]=[Ni2+]=3.42*10^-6 M

i) mass of Ni(OH)2=10.00 grams

volume of water=500 ml*(1L/1000ml)=0.5 L

As the solubility of Ni(OH)2 in aqueous medium (for saturated solution) =3.42*10^-6 M

So, mol of Ni(OH)2 dissolved in 0.5 L=(3.42*10^-6 mol/L)*0.5L=1.71*10^-6 mol

mass of Ni(OH)2 dissolved in 0.5 L=mol *molar mass=(1.71*10^-6 mol)*(92.708g/mol)=1.58*10^-4 g

mass of Ni(OH)2 undissolved at equilibrium=10.00 grams-1.58*10^-4 g=10.00 g (approx)

So, solid Ni(OH)2 remaining=10.00 grams (almost undissolved)

ii) Species in solution Ni2+(aq),OH-(aq),

[Ni2+]=S=3.42*10^-6 M

[OH-]=2*(3.42*10^-6 M)=6.84*10^-6 M

1b)

1) Ksp of Ba(OH)2=2.55*10^-4 =[Ba2+][OH-]^2

Let [Ba2+]=S''',[OH-]=S'' ,S''=solubility of Ba(OH)2

Ba(OH)2(s)<--->Ba2+ (aq) +2OH-(aq)

2.55*10^-4 =[Ba2+][OH-]^2=S''*(2S'')^2=4S''^3

S''=0.040M

mol of Ba(OH)2 dissolved in 0.5 L of water=0.040 mol/L*0.5 L=0.02 mol

mol of Ba(OH)2 dissolved in 0.5 L of water=0.02 mol*(171.34 g/mol)=3.42 grams

mass of solid Ba(OH)2=10.00-3.42 =6.58 grams

(As the solubility of Ba(OH)2 is much higher than that of Ni(OH)2 ,so Ba(OH)2 dissociates in the solution ,and the common ion OH- reverses the equilibrium of Ni(OH)2 so that its solubility further reduces)

mass of Ni(OH)2=10.00 grams(almost undissolved)

2) species in solution Ba2+,OH-

[Ba2+]=S''=0.040M

[OH-]=2*S''=2*0.040M=0.080M

3) Ni(OH)2 is slightly soluble in first problem but its solubility further reduces in the second problem due to common ion effect of OH- for both the salts at equilibrium.

As Ni(OH)2 has almost negligible solubility in water ,so the result is both the problem for Ni(OH)2 is approximately the same

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