To solve the problem of determining when a precipitate of AgCl is formed when equal volumes of AgNO3 and HCl are mixed, we will follow these steps:
### Step 1: Understand the Reaction
When silver nitrate (AgNO3) is mixed with hydrochloric acid (HCl), a precipitation reaction occurs, forming silver chloride (AgCl) and nitric acid (HNO3). The reaction can be represented as:
\[ \text{AgNO}_3 (aq) + \text{HCl} (aq) \rightarrow \text{AgCl} (s) + \text{HNO}_3 (aq) \]
### Step 2: Identify the Solubility Product (Ksp)
The solubility product constant (Ksp) for AgCl is given as:
\[ K_{sp} = 10^{-10} \]
### Step 3: Determine Ionic Product (Q)
The ionic product (Q) is calculated by multiplying the concentrations of the ions in the solution after mixing. If we assume the concentrations of AgNO3 and HCl are \( [Ag^+] \) and \( [Cl^-] \) respectively, and both solutions are mixed in equal volumes, the concentration of each ion after mixing will be halved.
For example, if the concentration of AgNO3 is \( C_{Ag} \) and the concentration of HCl is \( C_{Cl} \), then after mixing:
\[ [Ag^+] = \frac{C_{Ag}}{2} \]
\[ [Cl^-] = \frac{C_{Cl}}{2} \]
The ionic product \( Q \) is given by:
\[ Q = [Ag^+][Cl^-] = \left(\frac{C_{Ag}}{2}\right) \left(\frac{C_{Cl}}{2}\right) = \frac{C_{Ag} \cdot C_{Cl}}{4} \]
### Step 4: Compare Q with Ksp
To determine if a precipitate forms, we compare \( Q \) with \( K_{sp} \):
- If \( Q < K_{sp} \): No precipitate forms (solution is unsaturated).
- If \( Q = K_{sp} \): The solution is saturated, and a precipitate begins to form.
- If \( Q > K_{sp} \): A precipitate will form (solution is supersaturated).
### Step 5: Calculate Q for Given Concentrations
Now, we need to calculate \( Q \) for the provided concentrations in the options. For each option, we will compute \( Q \) and compare it to \( K_{sp} \).
1. **Option 1: \( C_{Ag} = 10^{-4} \, M \), \( C_{Cl} = 10^{-7} \, M \)**
\[
Q = \frac{10^{-4} \times 10^{-7}}{4} = \frac{10^{-11}}{4} = 2.5 \times 10^{-12} < 10^{-10} \quad (\text{No precipitate})
\]
2. **Option 2: \( C_{Ag} = 10^{-5} \, M \), \( C_{Cl} = 10^{-6} \, M \)**
\[
Q = \frac{10^{-5} \times 10^{-6}}{4} = \frac{10^{-11}}{4} = 2.5 \times 10^{-12} < 10^{-10} \quad (\text{No precipitate})
\]
3. **Option 3: \( C_{Ag} = 10^{-4} \, M \), \( C_{Cl} = 10^{-4} \, M \)**
\[
Q = \frac{10^{-4} \times 10^{-4}}{4} = \frac{10^{-8}}{4} = 2.5 \times 10^{-9} > 10^{-10} \quad (\text{Precipitate forms})
\]
4. **Option 4: \( C_{Ag} = 10^{-5} \, M \), \( C_{Cl} = 10^{-4} \, M \)**
\[
Q = \frac{10^{-5} \times 10^{-4}}{4} = \frac{10^{-9}}{4} = 2.5 \times 10^{-10} > 10^{-10} \quad (\text{Precipitate forms})
\]
### Conclusion
From the calculations, precipitates of AgCl will form in **Option 3** and **Option 4**.
