Question

Write a balanced net ionic equation to show why the solubility of Zn(CN)2(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid.

Answer 1

Step-by-step explanation:

The task above seeks to test your knowledge of solubility of ionic compounds.

Here are some factors that affect the solubility of ionic compounds;

• addition of an ion common to the equilibrium causes a decrease in the solubility of a sparingly soluble salt.

• pH

• formation of complex ions,

The knowledge of the concept above can help you to demonstrate

how these factors when properly manoeuvred can be used to isolate one metal ion from a mixture in solution.

To solve the task above here are the steps needed

1. Consider the dissolution of Zn(CN)_2(s).

Zn(CN)_2(s) ◀—–▶ Zn2+ + 2CN-

ksp =[Zn2+][CN-]² = 8.0 x 10-12.

In a saturated solution, Zn2+

and CN- are in equilibrium with solid Zn(CN)_2.

If a strong acid is added to the solution, the cyanide ion can react to form the Hydrogen cyanide and

water

2 CN- + 2H3O+ ◀——▶ 2 HCN + 2H2O

K = 1/Ka(HCN))

From the tables;

Ka of HCN = 6.2 × 10^ - 10.

K = (1/ka)² = (1/ 6.2 x 10^-10) = (1.6 × 10^-11)^2 = 2.6 ×10^-22.

Zn(CN)_2 contains a basic anion so its solubility increases in the presence of acid. Use Ksp[Zn(CN)_2] and Ka to

calculate the overall equilibrium constant for this reaction.

Adding these two equilibrium reactions shows the effect of adding a strong acid to a saturated Zn(CN)_2 solution

Answer 2

(1). Check Explanation.

(2). Kc = 8.0 × 10^-12.

Step-by-step explanation:

Step one: write out the balanced ionic dissociation equation.

Zn(CN)2(s) <======> Zn^2+ + 2CN^-.

The reaction above is an equilibrium Reaction and the ksp value= 8.0 x 10-12.

Step 2 : add Hydrogen ion(from the acid) to the cyanide ion, CN^-. [The cyanide ion is a Bronsted- Lowry's base and it has the tendency of accepting hydrogen ion to form HCN, which is a weak acid].

2 CN^- + 2H3O^+ <====> 2 HCN + 2H2O.

The ka value of HCN = 6.2 x 10^-10. Therefore, the K =( 1/ ka)^2 = 1/ 6.2 x 10^-10 = (1.6 × 10^-11)^2 = 2.6 ×10^-22.

ANSWER TO THE FIRST PART OF THE QUESTION ==> Zn(CN)2 is an ionic compound and it contains a basic anion therefore the solubilty increases in the presence of a strong acid.

ANSWER TO THE SECOND PART OF THE QUESTION : We will now add the equation in step one and step two together to give us the equilibrium constant,Kc.

Therefore, we now have;

Zn(CN)2(s) + 2H3O^+ <===> HCN + 2CN^- + H2O. Kc = 8.0 × 10^-12.