Question

2. For half-cell B, when NaOH(aq) is added to Cu2+(aq), a sparingly soluble solid, Cu(OH)2, will form.

a. Write the reaction for the formation of Cu(OH)2. The Ksp for Cu(OH)2 is 4.8 x 10^-20.

b. Interpret the magnitude of Ksp. When NaOH(aq) is added to Cu2+(aq), what will happen to the "free" [Cu2+]? Choose one.
[Cu2+] will increase
[Cu2+] will decrease
[Cu2+] will not change
cannot be predicted.

c. After the nearly insoluble solid forms, the next step in the lab has you add H2SO4(aq). Write the reaction that will occur.

d. When the H2SO4(aq) is added, predict what will happen to the "free" [Cu2+]? Choose one
[Cu2+] will increase
[Cu2+] will decrease
[Cu2+] will not change
cannot be predicted.

Answer 1

answer 1) the reaction between NaOH and Cu²⁺ is as follows

2NaOH (aq) + Cu²⁺ (aq) ------->   Cu(OH)₂ (s) + 2 Na⁺ (aq)

answer 2) Cu(OH)2   <====>   Cu²⁺    +   2OH^-  

                   s                                s             2s

Ksp =s (2s)²  

   48 * 10^-21 = 4 s³  

s = 12 * 10-7

solubility is = 1.2 * 10^-6 mol L^-1

the formation of Cu(OH)2 indicates that the solubility prosuct of CU2+ and OH- is lower as compared to its ionic product due to which precipitate forms..

the concentration of Cu2+ decreases .

3) CU(OH)₂ (s)    + H₂SO₄ (aq)   ------>    CuSO₄ (s)   + 2 H₂O (aq)

4) free [Cu2+] will not change

As H2SO4 attacks only CUSO4