Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).

The pH of 0.10 M is 4.49.

Explanation:

Given: Initial concentration of = 0.10 M

Let us assume that amount of dissociates is x. So, ICE table for dissociation of    is as follows.

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for value is as follows.

Hence,

Formula to calculate pH is as follows.

Substitute the values into above formula as follows.

Thus, we can conclude that the pH of 0.10 M is 4.49.

a c d and f

just finished the assignment

the answer for sodium is 0.186

the answer for rubidium is 0.251