The solubility of silver chloride (AgCl) is 1.9 * 10 ^ – 3 * g / L Calculate kop – Molar mass for silver chloride is 143.5 g/mol. Select one: a. 3.61 * 10 ^ – 9 * mol / L b. 4.68 * 10 ^ – 9 moll L c. 1.75 * 10 ^ – 10 mollL 0.6 * 0.34 * 10 ^ – 4 * mol / L
The Correct Answer and Explanation is :
To calculate the solubility product constant (KspK_{sp}) of silver chloride (AgCl), we need to follow several steps.
Given:
- Solubility of AgCl = 1.9×10−3 g/L1.9 \times 10^{-3} \, \text{g/L}
- Molar mass of AgCl = 143.5 g/mol
Step 1: Convert the solubility from grams per liter to moles per liter
The given solubility is in grams per liter, but we need it in moles per liter to calculate KspK_{sp}.
The molar mass of AgCl is 143.5 g/mol. To convert the solubility into moles per liter: Solubility (mol/L)=Solubility (g/L)Molar mass (g/mol)\text{Solubility (mol/L)} = \frac{\text{Solubility (g/L)}}{\text{Molar mass (g/mol)}} Solubility (mol/L)=1.9×10−3 g/L143.5 g/mol\text{Solubility (mol/L)} = \frac{1.9 \times 10^{-3} \, \text{g/L}}{143.5 \, \text{g/mol}} Solubility (mol/L)=1.32×10−5 mol/L\text{Solubility (mol/L)} = 1.32 \times 10^{-5} \, \text{mol/L}
Step 2: Set up the dissociation equation for AgCl
When silver chloride dissociates in water, it dissociates according to the following equation: AgCl (s)⇌Ag+(aq)+Cl−(aq)\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)
For every mole of AgCl that dissolves, it produces 1 mole of Ag+\text{Ag}^+ and 1 mole of Cl−\text{Cl}^-.
Step 3: Write the expression for KspK_{sp}
The solubility product constant KspK_{sp} is given by the product of the concentrations of the ions at equilibrium: Ksp=[Ag+][Cl−]K_{sp} = [\text{Ag}^+][\text{Cl}^-]
Since the concentrations of Ag+\text{Ag}^+ and Cl−\text{Cl}^- are both equal to the solubility of AgCl (1.32 x 10^-5 mol/L), we can substitute this value: Ksp=(1.32×10−5 mol/L)×(1.32×10−5 mol/L)K_{sp} = (1.32 \times 10^{-5} \, \text{mol/L}) \times (1.32 \times 10^{-5} \, \text{mol/L}) Ksp=1.74×10−10 mol2/L2K_{sp} = 1.74 \times 10^{-10} \, \text{mol}^2/\text{L}^2
Step 4: Compare with the answer choices
We find that the correct value for KspK_{sp} is approximately 1.75×10−10 mol2/L21.75 \times 10^{-10} \, \text{mol}^2/\text{L}^2, which matches option c.
Thus, the correct answer is c. 1.75×10−10 mol/L21.75 \times 10^{-10} \, \text{mol/L}^2.
Conclusion:
The solubility product constant for AgCl is calculated by determining its solubility in moles per liter and then applying the formula for the solubility product. This gives a value of Ksp=1.75×10−10K_{sp} = 1.75 \times 10^{-10}, which corresponds to option c.