To find the equilibrium constant ($K_{\text{sp}}$) for the dissolution of AgBr, we use the solubility product constant expression:

$\text{AgBr (s)} \leftrightarrow \text{Ag}^+ (\text{aq}) + \text{Br}^- (\text{aq})$

Given dataTo find the equilibrium constant ($K_{\text{sp}}$) for the dissolution of AgBr, we use the solubility product constant expression:

$\text{AgBr (s)} \leftrightarrow \text{Ag}^+ (\text{aq}) + \text{Br}^- (\text{aq})$

Given

• The solubility of AgBr is 0.138 mg in 10.0 L of water.
• Molar mass of AgBr is approximately 187.77 g/mol.

Step 1: Convert solubility to molarity

First, convert the mass from mg to grams: $0.138 \text{ mg} = 0.138 \times 10^{-3} \text{ g} = 0.000138 \text{ g}$

Now, convert this mass to moles using the molar mass: $\text{Moles of AgBr} = \frac{0.000138 \text{ g}}{187.77 \text{ g/mol}} \approx 7.35 \times 10^{-7} \text{ mol}$

Since this amount is dissolved in 10.0 L of water, the molarity ($[ \text{Ag}^+ ] = [ \text{Br}^- ]$) is: $\text{Molarity} = \frac{7.35 \times 10^{-7} \text{ mol}}{10.0 \text{ L}} = 7.35 \times 10^{-8} \text{ M}$

Step 2: Calculate the equilibrium constant ($K_{\text{sp}}$)

The solubility product constant is given by: [ K_{\text{sp}} = [ \text{Ag}^+ ][ \text{Br}^- ] = (7.35 \times 10^{-8})(7.35 \times 10^{-8}) = 5.40 \times 10^{-15} ]

Thus, the equilibrium constant for the dissolution of AgBr is approximately $5.40 \times 10^{-15}$.

Would you like further details or have any questions?

Here are some related questions:

1. How would increasing the volume of water affect the solubility of AgBr?
2. What is the effect of adding common ions like Ag+ or Br- to the solution?
3. How can the solubility of AgBr change with temperature?
4. What is the relationship between Ksp and the Gibbs free energy for a dissolution process?
5. How does the ionic strength of a solution affect the activity coefficients in the calculation of Ksp?

Tip: The solubility product constant (Ksp) can be used to predict whether a precipitate will form when two solutions are mixed.