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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.

At 25°C, 0.138 mg AgBr dissolves in 10.0 L of water. What is the equilibrium constant for the reaction below? AgBr(s) ⇌ Ag+(aq) + Br−(aq)

Learn how to calculate the equilibrium constant (Ksp) for AgBr dissolving in water at 25°C. This problem involves solubility product constants, stoichiometry, and molarity calculations. Suitable for high school chemistry students.

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