To solve the problem, we need to determine the concentration of \( \text{Cu}^{2+} \) ions after electrolysis when an electric current is passed through a solution of blue vitriol (copper sulfate) and silver nitrate. Here’s a step-by-step breakdown of the solution:
### Step 1: Determine the weight of copper deposited
From the problem, we know that 1.307 g of silver was deposited during electrolysis. We will use Faraday's second law of electrolysis to find the weight of copper deposited.
### Step 2: Use Faraday's second law of electrolysis
According to Faraday's second law:
\[
\frac{\text{Weight of Cu deposited}}{\text{Weight of Ag deposited}} = \frac{Z_{\text{Cu}}}{Z_{\text{Ag}}}
\]
Where \( Z \) is the electrochemical equivalent.
### Step 3: Calculate the electrochemical equivalents
The electrochemical equivalent \( Z \) can be calculated using:
\[
Z = \frac{\text{Molar mass}}{n} \times \frac{1}{96500}
\]
For silver (\( \text{Ag} \)):
- Molar mass of \( \text{Ag} = 108 \, \text{g/mol} \)
- \( n \) (number of electrons) for silver = 1
\[
Z_{\text{Ag}} = \frac{108}{1} \times \frac{1}{96500} = 0.00112 \, \text{g/C}
\]
For copper (\( \text{Cu} \)):
- Molar mass of \( \text{Cu} = 63.5 \, \text{g/mol} \)
- \( n \) for copper = 2
\[
Z_{\text{Cu}} = \frac{63.5}{2} \times \frac{1}{96500} = 0.000329 \, \text{g/C}
\]
### Step 4: Set up the equation
Substituting the values into the equation:
\[
\frac{\text{Weight of Cu}}{1.307} = \frac{0.000329}{0.00112}
\]
Let \( \text{Weight of Cu} = x \):
\[
\frac{x}{1.307} = \frac{0.000329}{0.00112}
\]
Cross-multiplying gives:
\[
x = 1.307 \times \frac{0.000329}{0.00112}
\]
Calculating \( x \):
\[
x \approx 0.384 \, \text{g}
\]
### Step 5: Calculate the initial moles of \( \text{Cu}^{2+} \)
The initial weight of blue vitriol (copper sulfate) is 10 g in 500 ml of water. The molar mass of copper sulfate pentahydrate \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is approximately 249.5 g/mol.
Calculating the number of moles of \( \text{CuSO}_4 \):
\[
\text{Moles of CuSO}_4 = \frac{10 \, \text{g}}{249.5 \, \text{g/mol}} \approx 0.0400 \, \text{mol}
\]
Since each mole of copper sulfate produces one mole of \( \text{Cu}^{2+} \):
\[
\text{Moles of } \text{Cu}^{2+} = 0.0400 \, \text{mol}
\]
### Step 6: Calculate the moles of \( \text{Cu}^{2+} \) after electrolysis
The weight of copper deposited is 0.384 g. To find the moles of copper deposited:
\[
\text{Moles of Cu} = \frac{0.384 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.00604 \, \text{mol}
\]
Now, subtract the moles of copper deposited from the initial moles:
\[
\text{Remaining moles of } \text{Cu}^{2+} = 0.0400 - 0.00604 = 0.03396 \, \text{mol}
\]
### Step 7: Calculate the concentration of \( \text{Cu}^{2+} \)
To find the concentration (molarity) of \( \text{Cu}^{2+} \):
\[
\text{Molarity} = \frac{\text{Moles of } \text{Cu}^{2+}}{\text{Volume in L}} = \frac{0.03396 \, \text{mol}}{0.5 \, \text{L}} \approx 0.06792 \, \text{mol/L} \approx 0.068 \, \text{mol/L}
\]
### Final Answer
The concentration of \( \text{Cu}^{2+} \) after electrolysis is approximately **0.068 mol/L**.
