i. Relation between pH and pOH at 25°C:

At 25°C, the relationship between pH and pOH in an aqueous solution is derived from the ionization of water:

$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$

The ionization constant of water ($K_w$) is given by:

[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \quad \text{(at 25°C)} ]

The pH and pOH are defined as:

$\text{pH} = -\log [\text{H}^+]$ $\text{pOH} = -\log [\text{OH}^-]$

Taking the logarithm of $K_w$:

[ -\log K_w = -\log [\text{H}^+] - \log [\text{OH}^-] ]

Thus:

$\text{pH} + \text{pOH} = 14$

So, at 25°C, the sum of pH and pOH of any aqueous solution is always equal to 14.

ii. Solubility Product and Mass of Precipitated AgCl:

Given:

• Solubility product ($K_{sp}$) of AgCl = $1.8 \times 10^{-10}$
• Molecular mass of AgCl = 143.5 g/mol
• Concentration of HCl = 0.2 M

First, determine the concentration of Cl⁻ ions contributed by the added HCl:

$[\text{Cl}^-] = 0.2 \, \text{M}$

The solubility product expression for AgCl is:

[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] ]

At saturation, if the concentration of Ag⁺ is $s$:

$K_{sp} = s \times [\text{Cl}^-]$

Substituting the values:

$1.8 \times 10^{-10} = s \times 0.2$

Solving for $s$:

$s = \frac{1.8 \times 10^{-10}}{0.2} = 9 \times 10^{-10} \, \text{M}$

This represents the concentration of Ag⁺ ions remaining in solution after the precipitation of AgCl. Now, calculate the initial concentration of Ag⁺ ions in the saturated solution of AgCl (before adding HCl):

$s_0 = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5} \, \text{M}$

The change in Ag⁺ concentration due to precipitation is:

$\Delta s = s_0 - s \approx 1.34 \times 10^{-5} - 9 \times 10^{-10} \approx 1.34 \times 10^{-5} \, \text{M}$

This is the concentration of Ag⁺ ions precipitated as AgCl.

Now, calculate the mass of AgCl precipitated:

$\text{Mass} = \Delta s \times \text{Volume} \times \text{Molecular mass of AgCl}$

Assuming 1 liter of solution:

$\text{Mass} = 1.34 \times 10^{-5} \, \text{mol/L} \times 1 \, \text{L} \times 143.5 \, \text{g/mol} \approx 1.92 \times 10^{-3} \, \text{g}$

Thus, approximately 1.92 mg of AgCl will precipitate when 0.2 moles of HCl is added to the saturated solution of AgCl.

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Would you like more details on these calculations, or do you have any questions?

Here are some related questions you might find useful:

1. What is the relationship between $K_w$ and temperature?
2. How does the addition of a common ion affect solubility?
3. What is the concept of ionic strength in relation to solubility?
4. Can the solubility product change with temperature?
5. How do you calculate the pH of a solution when both H⁺ and OH⁻ ions are present?

Tip: Remember that the solubility of a salt decreases when a common ion is added to the solution due to the common ion effect.