To solve the problem of how much `Ag^+` remains in solution after mixing with `HCl`, we can follow these steps:
### Step 1: Calculate the initial concentration of `Ag^+` ions
We are given that there are `10^(-3)` moles of `Ag^+` in `50 mL` of solution. To find the concentration, we use the formula:
\[
\text{Concentration of } Ag^+ = \frac{\text{Number of moles}}{\text{Volume in liters}}
\]
Convert `50 mL` to liters:
\[
50 \, \text{mL} = 0.050 \, \text{L}
\]
Now, calculate the concentration:
\[
\text{Concentration of } Ag^+ = \frac{10^{-3} \, \text{moles}}{0.050 \, \text{L}} = 0.02 \, \text{M}
\]
### Step 2: Calculate the concentration of `Cl^-` ions from `HCl`
We have `50 mL` of `0.1 M HCl`. The number of moles of `Cl^-` ions is:
\[
\text{Moles of } Cl^- = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles}
\]
After mixing, the total volume becomes `100 mL` or `0.1 L`. The concentration of `Cl^-` ions in the total volume is:
\[
\text{Concentration of } Cl^- = \frac{0.005 \, \text{moles}}{0.1 \, \text{L}} = 0.05 \, \text{M}
\]
### Step 3: Calculate the ionic product (Q) for `AgCl`
The ionic product `Q` for the precipitation of `AgCl` is given by:
\[
Q = [Ag^+][Cl^-]
\]
Substituting the concentrations we found:
\[
Q = (0.02)(0.05) = 0.001 = 1 \times 10^{-3}
\]
### Step 4: Compare `Q` with `Ksp` of `AgCl`
The solubility product constant \( K_{sp} \) of `AgCl` is given as \( 1.0 \times 10^{-10} \). Since \( Q > K_{sp} \), this indicates that a precipitate will form.
### Step 5: Determine the amount of `Ag^+` that precipitates
When `Ag^+` and `Cl^-` react to form `AgCl`, they will react in a 1:1 ratio. Initially, we have:
- `Ag^+` = `0.02 moles`
- `Cl^-` = `0.005 moles`
Since `Cl^-` is the limiting reactant, all `Cl^-` will react with `Ag^+`. Therefore, `0.005 moles` of `Ag^+` will react, leaving:
\[
\text{Remaining } Ag^+ = 0.02 - 0.005 = 0.015 \, \text{moles}
\]
### Step 6: Calculate the concentration of remaining `Ag^+` after precipitation
After precipitation, the total volume of the solution is still `100 mL` or `0.1 L`. Therefore, the concentration of remaining `Ag^+` is:
\[
\text{Concentration of remaining } Ag^+ = \frac{0.015 \, \text{moles}}{0.1 \, \text{L}} = 0.15 \, \text{M}
\]
### Step 7: Use the `Ksp` to find the equilibrium concentration of `Ag^+`
At equilibrium, the concentration of `Ag^+` can be calculated using the `Ksp` expression:
\[
K_{sp} = [Ag^+][Cl^-]
\]
Let \( x \) be the concentration of `Ag^+` remaining in solution after precipitation. The concentration of `Cl^-` after precipitation will be:
\[
[Cl^-] = 0.05 - 0.005 = 0.045 \, \text{M}
\]
Substituting into the `Ksp` expression:
\[
1.0 \times 10^{-10} = x \cdot 0.045
\]
Solving for \( x \):
\[
x = \frac{1.0 \times 10^{-10}}{0.045} \approx 2.22 \times 10^{-9} \, \text{M}
\]
### Conclusion
The concentration of `Ag^+` remaining in solution after precipitation is approximately \( 2.22 \times 10^{-9} \, \text{M} \).
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