Will a precipitate be observed if 0.10 mol Ag* and 0.001 mol SO4²- are added to make 1.0 L of solution? (Ksp = 1.4 x 10-5) yes, because Q < Ksp no, because Q > Ksp no, because Q < Ksp

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**Precipitation Chemistry Problem**

**Question:**
Will a precipitate be observed if 0.10 mol Ag⁺ and 0.001 mol SO₄²⁻ are added to make 1.0 L of solution? (Ksp = 1.4 × 10⁻⁵)

**Options:**
- ○ yes, because Q < Ksp

- ○ no, because Q > Ksp

- ○ no, because Q < Ksp

When attempting to determine if a precipitate will form in a solution, it is essential to compare the ion product (Q) with the solubility product constant (Ksp). If Q > Ksp, a precipitate will form because the solution is supersaturated. If Q < Ksp, no precipitate will form because the solution is unsaturated. If Q = Ksp, the solution is at equilibrium and is saturated.

**Calculation Process:**
To solve this, first, determine the molar concentrations of Ag⁺ and SO₄²⁻ in the solution. Since the volume of the solution is 1.0 L:

\[ [Ag^+] = \frac{0.10 \text{ mol}}{1.0 \text{ L}} = 0.10 \text{ M} \]
\[ [SO₄²⁻] = \frac{0.001 \text{ mol}}{1.0 \text{ L}} = 0.001 \text{ M} \]

Next, calculate the reaction quotient (Q) for the precipitation reaction:

\[ Q = [Ag^+] \times [SO₄²⁻] \]
\[ Q = (0.10 \text{ M}) \times (0.001 \text{ M}) \]
\[ Q = 1.0 \times 10^{-4} \]

Compare Q with the given Ksp:

\[ Q = 1.0 \times 10^{-4} \]
\[ Ksp = 1.4 \times 10^{-5} \]

Since \( Q > Ksp \), the solution is supersaturated, and a precipitate will form. Therefore, the correct option is:

- ○ no, because Q > Ksp
Transcribed Image Text:**Precipitation Chemistry Problem** **Question:** Will a precipitate be observed if 0.10 mol Ag⁺ and 0.001 mol SO₄²⁻ are added to make 1.0 L of solution? (Ksp = 1.4 × 10⁻⁵) **Options:** - ○ yes, because Q < Ksp - ○ no, because Q > Ksp - ○ no, because Q < Ksp When attempting to determine if a precipitate will form in a solution, it is essential to compare the ion product (Q) with the solubility product constant (Ksp). If Q > Ksp, a precipitate will form because the solution is supersaturated. If Q < Ksp, no precipitate will form because the solution is unsaturated. If Q = Ksp, the solution is at equilibrium and is saturated. **Calculation Process:** To solve this, first, determine the molar concentrations of Ag⁺ and SO₄²⁻ in the solution. Since the volume of the solution is 1.0 L: \[ [Ag^+] = \frac{0.10 \text{ mol}}{1.0 \text{ L}} = 0.10 \text{ M} \] \[ [SO₄²⁻] = \frac{0.001 \text{ mol}}{1.0 \text{ L}} = 0.001 \text{ M} \] Next, calculate the reaction quotient (Q) for the precipitation reaction: \[ Q = [Ag^+] \times [SO₄²⁻] \] \[ Q = (0.10 \text{ M}) \times (0.001 \text{ M}) \] \[ Q = 1.0 \times 10^{-4} \] Compare Q with the given Ksp: \[ Q = 1.0 \times 10^{-4} \] \[ Ksp = 1.4 \times 10^{-5} \] Since \( Q > Ksp \), the solution is supersaturated, and a precipitate will form. Therefore, the correct option is: - ○ no, because Q > Ksp
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