Predict whether a precipitate will be formed or not on mixing 20 mL of 0.001 M NaCI with 80 mL of 0.01 M `AgNO_(3)` solution `(K_(sp) " for " AgCI =1.5 xx 10^(-10))`

Explanation

To determine whether a precipitate will form when mixing 20 mL of 0.001 M NaCl with 80 mL of 0.01 M AgNO3, we can follow these steps:

### Step 1: Calculate the moles of NaCl and AgNO3 before mixing.
1. **Moles of NaCl**:
\[ \text{Moles of NaCl} = \text{Concentration} \times \text{Volume} = 0.001 \, \text{M} \times 0.020 \, \text{L} = 0.00002 \, \text{moles} \]

2. **Moles of AgNO3**:
\[ \text{Moles of AgNO3} = \text{Concentration} \times \text{Volume} = 0.01 \, \text{M} \times 0.080 \, \text{L} = 0.0008 \, \text{moles} \]

### Step 2: Calculate the total volume after mixing.
\[ \text{Total Volume} = 20 \, \text{mL} + 80 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

### Step 3: Calculate the concentrations of Na+ and Cl- ions after mixing.
1. **Concentration of Na+**:
\[ [\text{Na}^+] = \frac{\text{Moles of NaCl}}{\text{Total Volume}} = \frac{0.00002 \, \text{moles}}{0.1 \, \text{L}} = 0.0002 \, \text{M} \]

2. **Concentration of Cl-**:
\[ [\text{Cl}^-] = \frac{\text{Moles of NaCl}}{\text{Total Volume}} = \frac{0.00002 \, \text{moles}}{0.1 \, \text{L}} = 0.0002 \, \text{M} \]

### Step 4: Calculate the concentration of Ag+ ions after mixing.
1. **Concentration of Ag+**:
\[ [\text{Ag}^+] = \frac{\text{Moles of AgNO3}}{\text{Total Volume}} = \frac{0.0008 \, \text{moles}}{0.1 \, \text{L}} = 0.008 \, \text{M} \]

### Step 5: Calculate the ion product (Q) for AgCl.
\[ Q = [\text{Ag}^+][\text{Cl}^-] = (0.008 \, \text{M})(0.0002 \, \text{M}) = 1.6 \times 10^{-6} \]

### Step 6: Compare Q with Ksp.
- Given \( K_{sp} \) for AgCl = \( 1.5 \times 10^{-10} \)

Since \( Q (1.6 \times 10^{-6}) > K_{sp} (1.5 \times 10^{-10}) \), a precipitate of AgCl will form.

### Final Conclusion:
A precipitate of AgCl will be formed when mixing the two solutions.

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