Copper Ion Equilibrium Solution color and clarity (clear or cloudy) of copper (I

Question

Copper Ion Equilibrium

Solution color and clarity (clear or cloudy) of copper (II) nitrate solution.

Color: Blue

Clarity: Clear

Solution color and clarity after addition of 1M NaOH. (Step2)

Color: Light blue

Clarity: Cloudy

Solution color and clarity after the addition of 3M HCl (Step3) and comparison of solutions from steps 2 and 3.

Color: Blue

Clarity: Cloudy

Describe in terms of equilibrium steps 2 and 3 by completing the model explanation: Adding the source of OH-­ in step 2 pushed equilibrium a) to the LEFT or RIGHT (circle one). This made the solid Cu(OH)2 APPEAR or DISAPPEAR (circle one). The H+ added in step 3 turns some of the OH-­ to water thereby removing some of the hydroxide ions from solution. This causes equilibrium a) to shift LEFT or RIGHT and the solid to APPEAR or DISAPPEAR.

Observations from addition of NaOH and HCl.

-Addition of NaOH causes the solution to become cloudy and light blue

-Addition of HCl causes the solution to be blue and clear.

Solution color and clarity in second test tube with copper (II) nitrate and ammonia. What substance do you apparently have here?

Color: Light blue

Clarity: Cloudy

Substance: ??????

a)Cu(OH)2(s) <--> Cu2+(aq) + 2OH-­(aq)

b)NH3(aq) + H2O(l) <--> NH4+(aq) + OH-­(aq)

c)Cu(NH3)42+(aq) <--> Cu2+(aq) + 4NH3(aq)

Explain your observations above in terms of equilibrium a) and b). DO NOT try to combine a) and b) into one equation; use them separately, but point out how they affect one another: ????

Solution color and clarity after addition of 3M HCl.

Color: Blue

Clarity: Clear

Explanation of observations using equilibrium a): ?????

Complete description of solution after addition of NH3. What is the final color?

Color: Dark blue

              Clarity: Clear

Explanation of observations in terms of equilibria a), b) and c): ????

Observations of solution after addition of HCl.

Color: Blue

Clarity: Clear

Explanation of observations in terms of equilibria a), b) and c): ?????

Please answer the ????? parts.

Explanation / Answer

first test tube

When 1M NaOH solution was added to copper (II) nitrate solution. the following reaction took place:

Cu(NO3)2(aq) + 2 NaOH(aq) --> Cu(OH)2(s) + 2 NaNO3(aq)

Adding the source of OH- in step 2 pushed equilibrium a) to the RIGHT. This made the solid Cu(OH)2 APPEAR.

The H+ added in step 3 turns some of the OH-­ to water thereby removing some of the hydroxide ions from solution. The system wuold shift to the left and the solid Cu(OH)2 to DISAPPEAR

Cu(OH)2 + 2HCl CuCl2 + 2H2O

second test tube

When ammonia (NH3) is added to copper(II) nitrate (Cu(NO3)2), the result is a pale blue precipitate, which is Cu(OH)2(s). this is the substance.

a)Cu(OH)2(s) <--> Cu2+(aq) + 2OH-­(aq)

saturated solution of an ionic compound has both ions in solution and ions in the solid. The mixture may appear cloudy (the solid suspended in the solution) or the solid may settle to the bottom.The concentration of ions in the solution has reached a maximum. If more solid is added the concentration remains constant. The ions in the solid and solution are in equilibrium.The ions are simultaneously dissociating into solution (forward reaction) and crystallizing into the solid (reverse reaction). Both reactions are still occurring. In this dynamic equilibrium the rate of the forward reaction equals the rate of the reverse reaction.

b)NH3(aq) + H2O(l) <--> NH4+(aq) + OH-­(aq)

The equilibrium constant for this reaction is written as follows:

Kc = ( [NH4+] [OH¯] ) / ( [NH3] [H2O] )

However, in pure liquid water, [H2O] is a constant value. Moving [H2O] to the other side gives:

Kc [H2O] = ( [NH4+] [OH¯] ) / [NH3]

Since the term Kc [H2O] is a constant, let it be symbolized by Kb, giving:

Kb = ( [NH4+] [OH¯] ) / [NH3]

This constant, Kb, is called the base ionization constant.

c)Cu(NH3)42+(aq) <--> Cu2+(aq) + 4NH3(aq)

how a) and b) affect one another?

In the presence of excess NH3, the equilibrium of a) shifts to the left as the free Cu 2+ ion is sequestered from the solution as the Cu(NH3)42+ complex ion.    

Cu 2+ (aq) + 4 NH3 (aq) <=> Cu(NH3)42+ (aq)  

Observations of solution after addition of HCl:

The H+ added in step 3 turns some of the OH- to water thereby removing some of the hydroxide ions from solution. The system wuold shift to the left and the solid Cu(OH)2 to DISAPPEAR

Cu(OH)2 + 2HCl CuCl2 + 2H2O

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