1. 0.08M HOCN [Given that :Ksp for AgOCN =2.3 x 10^-7 ; Ka(HOCN) = 3.3 x 10^-4 ]

To analyze the solubility of silver cyanate (AgOCN) in a solution of 0.08 M HOCN, we need to consider both the solubility product constant (Ksp) of AgOCN and the acid dissociation constant (Ka) of HOCN. This situation involves a common ion effect due to the presence of cyanate ions (OCN^-) from HOCN, which can influence the solubility of AgOCN.

Understanding the Components

First, let's break down the relevant constants:

• Ksp for AgOCN: This is the solubility product constant, which tells us the equilibrium between the solid and its ions in solution. For AgOCN, the dissociation can be represented as:
• AgOCN (s) ⇌ Ag⁺ (aq) + OCN^- (aq)
• Ka for HOCN: This constant indicates the strength of the weak acid HOCN in water:
• HOCN (aq) ⇌ H⁺ (aq) + OCN^- (aq)

Setting Up the Equilibrium

When AgOCN is added to the 0.08 M HOCN solution, the dissociation of HOCN produces OCN^- ions, which will shift the equilibrium of AgOCN's dissociation due to the common ion effect. This means that the presence of OCN^- will decrease the solubility of AgOCN.

Calculating the Solubility of AgOCN

Let’s denote the solubility of AgOCN in the 0.08 M HOCN solution as 's'. The equilibrium concentrations can be expressed as follows:

• [Ag⁺] = s
• [OCN^-] = s + 0.08

Now, we can write the expression for Ksp:

Ksp = [Ag⁺][OCN^-]

Substituting the equilibrium concentrations into the Ksp expression gives:

Ksp = s(s + 0.08)

Given that Ksp for AgOCN is 2.3 x 10^-7, we can set up the equation:

2.3 x 10^-7 = s(s + 0.08)

Solving the Equation

This is a quadratic equation in the form of:

s² + 0.08s - 2.3 x 10^-7 = 0

We can apply the quadratic formula, s = [-b ± √(b² - 4ac)] / 2a, where:

• a = 1
• b = 0.08
• c = -2.3 x 10^-7

Calculating the discriminant:

b² - 4ac = (0.08)² - 4(1)(-2.3 x 10^-7)

After calculating, we find the roots for s, which will give us the solubility of AgOCN in the presence of HOCN.

Final Considerations

Once you find the value of 's', it will represent the solubility of AgOCN in the 0.08 M HOCN solution. This example illustrates how the common ion effect can significantly alter the solubility of a salt in solution, a key concept in chemical equilibria.