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Equilibrium / SolubilityFRQ

Cu(OH)2(s)  Cu2+ (aq) + 2OH– (aq) Ksp = 2.2  10–20 Cu 2+(aq) + 4NH3(aq)  Cu(NH3)4 2+(aq) Kf = 2.1Equilibrium / Solubility Chemistry Question

Problem Context

Cu(OH)2(s)  Cu2+ (aq) + 2OH– (aq) Ksp = 2.2  10–20
Cu 2+(aq) + 4NH3(aq)  Cu(NH3)4 2+(aq) Kf = 2.1  1013
Use the equations and K values above to answer the questions below.

a.

Determine the solubility of Cu(OH)2 in mol/L at pH = 8.00

Model Answer

Cu(OH)2 Cu2+ + 2 OH– pH = 8 pOH = 6
ksp = [Cu2+] × [OH–]2
2.2 × 10–20 = [Cu2+] × [10–6]2
[Cu2+] = 2.2 × 10–8 M

b.

If 20.0 mL of 0.0010 M CuSO4 is mixed with 50.0 mL of 0.0010 M NaOH, determine whether Cu(OH)2 will precipitate. Support your answer with appropriate calculations.

Model Answer

= 2.86 × 10–4 M Cu2+

= 7.14 × 10–4 M OH–
Q = (2.86 × 10–4) × (7.14 × 10–4) 2
Q = 1.46 × 10–10 >> Ksp ∴ Ppt

c.

Write the equation for the reaction of Cu(OH)2 with aqueous NH3 and calculate the K value for this reaction.

Model Answer

Cu(OH)2 + 4 NH3  Cu(NH3)42+ + 2 OH–
K = Ksp × Kf
K = (2.2 × 10–20) × (2.1 × 1013) = 4.6 × 10–7

d.

Calculate the [NH3] that would be needed to dissolve 0.100 g of Cu(OH)2 in 1.00 L of H2O.

Model Answer

1 g Cu(OH)2 ×
= 0.00103 mol Cu(OH)2 in 1 L.
OH– will be 0.00206 M

4.6 × 10–7 = [Cu(NH3)4] × (OH–)2 = 0.00103 × (0.00206)2
[NH3]4 = 0.00950
[NH3] = (0.00950)1/4
[NH3] = 0.312 M

e.

Describe what would be observed if 5.0 M NH3 is added dropwise to a 0.10 M solution of Cu 2+ ions.

Model Answer

As the NH3 solution is added a blue precipitate will form which will dissolve as more NH3 is added to form a dark blue solution.

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