the solubility of Pb(OH)₂ in water is 5*10^-6 u. Calculate the ph of buffer solution in which solubility of Pb(OH)₂ is 0.5*10^-3 mol/litre

To calculate the pH of a buffer solution in which the solubility of lead(II) hydroxide, Pb(OH)₂, is 0.5 x 10⁻³ mol/L, we first need to understand the relationship between the solubility of Pb(OH)₂ and the concentration of hydroxide ions (OH⁻) it produces in solution. Lead(II) hydroxide is a sparingly soluble compound, and its solubility product constant (Ksp) can help us determine the pH of the solution.

Understanding the Dissolution of Pb(OH)₂

When Pb(OH)₂ dissolves in water, it dissociates according to the following equation:

• Pb(OH)₂ (s) ⇌ Pb²⁺ (aq) + 2 OH⁻ (aq)

From this equation, we can see that for every mole of Pb(OH)₂ that dissolves, it produces one mole of Pb²⁺ ions and two moles of hydroxide ions. Therefore, if the solubility of Pb(OH)₂ is 0.5 x 10⁻³ mol/L, the concentration of OH⁻ ions in the solution will be:

• [OH⁻] = 2 × (0.5 x 10⁻³) = 1.0 x 10⁻³ mol/L

Calculating the pOH

Next, we can calculate the pOH of the solution using the formula:

• pOH = -log[OH⁻]

Substituting the concentration of hydroxide ions:

• pOH = -log(1.0 x 10⁻³) = 3

Finding the pH

To find the pH of the solution, we can use the relationship between pH and pOH:

• pH + pOH = 14

Now, substituting the value of pOH we just calculated:

• pH = 14 - pOH = 14 - 3 = 11

Final Thoughts

Thus, the pH of the buffer solution in which the solubility of Pb(OH)₂ is 0.5 x 10⁻³ mol/L is 11. This indicates a basic solution, which is consistent with the presence of hydroxide ions from the dissolution of lead(II) hydroxide. Understanding these relationships helps in predicting the behavior of sparingly soluble salts in various solutions.