7) (7 points) A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 5

Question

7) (7 points) A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mixture. What concentration of hydroxide is required to precipitate each metal? Put a star next to the metal that precipitates first. (Ksp = 2.2 x 10–20 for Cu(OH)2, Ksp = 1.3 x 10–15 for Co(OH)2) [OH-] to precipitate Cu2+: ______________ [OH-] to precipitate Co2+: ______________ 8) (7 points) Barbital (MW = 184.2 g/mol) is a simple barbiturate used for sedation. What is the boiling point of a solution prepared by dissolving 42.5 g of barbital, a non-electrolyte, in 825 g of acetic acid? (Kb = 3.07 °C/m; boiling point of pure acetic acid = 117.9°C)

Explanation / Answer

7)

Ksp of Cu(OH)2 = 2.2 x 10^-20

Ksp of Co(OH)2 = 1.3 x 10^-15

Cu(OH)2 will precipitates first

Cu(OH)2   ------------> Cu+2   +    2 OH-

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

2.2 x 10^-20 = (0.25) x [OH-]^2

precipitate Cu+2 [OH-] = 2.97 x 10^-10 M

Co(OH)2   --------------> Co+2   +   2 OH-

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

1.3 x 10^-15 = 0.25 x [OH-]^2

precipitate Co+2 [OH-] = 7.2 x 10^-8 M

8)

moles = 42.5 / 184.2 = 0.2307

molality = 0.2307 / 0.825 = 0.2797 m

delta Tb = Kb x m

Tb - To = Kb x m

Tb - 117.9 = 3.07 x 0.2797

Tb = 118.76 oC

boiling point = 118.76 oC

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