450 mL of 0.001 N solution of `AgNO_3` is added to 50 mL of 0.001 N solution of HCI. Will there be a formation of precipitate of AgCl? `K_(sp) (AgCl)=1.8 xx 10^(-10)`.

Answer

To determine whether a precipitate of AgCl will form when 450 mL of 0.001 N AgNO3 is added to 50 mL of 0.001 N HCl, we can follow these steps:

### Step 1: Calculate the moles of AgNO3 and HCl
1. **Calculate the moles of AgNO3:**
- Volume of AgNO3 solution = 450 mL = 0.450 L
- Normality of AgNO3 = 0.001 N
- Moles of AgNO3 = Normality × Volume (in L) = 0.001 N × 0.450 L = 0.00045 moles

2. **Calculate the moles of HCl:**
- Volume of HCl solution = 50 mL = 0.050 L
- Normality of HCl = 0.001 N
- Moles of HCl = Normality × Volume (in L) = 0.001 N × 0.050 L = 0.00005 moles

### Step 2: Determine the concentrations of Ag⁺ and Cl⁻ after mixing
1. **Total volume after mixing:**
- Total volume = 450 mL + 50 mL = 500 mL = 0.500 L

2. **Concentration of Ag⁺ ions:**
- Moles of Ag⁺ = Moles of AgNO3 = 0.00045 moles
- Concentration of Ag⁺ = Moles / Total Volume = 0.00045 moles / 0.500 L = 0.0009 M = 9 × 10^(-4) M

3. **Concentration of Cl⁻ ions:**
- Moles of Cl⁻ = Moles of HCl = 0.00005 moles
- Concentration of Cl⁻ = Moles / Total Volume = 0.00005 moles / 0.500 L = 0.0001 M = 1 × 10^(-4) M

### Step 3: Calculate the ionic product (IP) of AgCl
- The ionic product (IP) of AgCl is given by the formula:
\[ IP = [Ag^+] \times [Cl^-] \]
- Substituting the concentrations:
\[ IP = (9 \times 10^{-4}) \times (1 \times 10^{-4}) = 9 \times 10^{-8} \]

### Step 4: Compare the ionic product with Ksp
- Given \( K_{sp} \) of AgCl = \( 1.8 \times 10^{-10} \)
- Compare IP with Ksp:
- \( IP = 9 \times 10^{-8} \)
- \( K_{sp} = 1.8 \times 10^{-10} \)

Since \( IP > K_{sp} \), a precipitate will form.

### Conclusion
Yes, a precipitate of AgCl will form when 450 mL of 0.001 N AgNO3 is added to 50 mL of 0.001 N HCl.

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