When equal volumes of the following solutions are mixed, precipitation of AgCl `(K_(sp)=1.8 xx 10^(-10))` will occur only with
(a) `10^(-4) M (Ag^(+)) and 10^(-4) M (Cl^(-))`
(b) `10^(-5) M (Ag^(+)) and 10^(-5) M (Cl^(-))`
(c) `10^(-6) M (Ag^(+)) and 10^(-6) M (Cl^(-))`
(d) `10^(-10) M (Ag^(+)) and 10^(-10) M (Cl^(-))`

Answer

To determine when precipitation of AgCl will occur when equal volumes of the given solutions are mixed, we can follow these steps:

### Step 1: Understand the concept of Ksp
The solubility product constant (Ksp) for AgCl is given as \( K_{sp} = 1.8 \times 10^{-10} \). Precipitation occurs when the ionic product (IP) of the ions exceeds the Ksp value.

### Step 2: Calculate the concentrations after mixing
When equal volumes of two solutions are mixed, the concentration of each ion will be halved.

### Step 3: Analyze each option

**Option (a): \( 10^{-4} M (Ag^{+}) \) and \( 10^{-4} M (Cl^{-}) \)**
- After mixing:
\[ [Ag^{+}] = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, M \]
\[ [Cl^{-}] = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, M \]
- Calculate the ionic product (IP):
\[ IP = [Ag^{+}][Cl^{-}] = (5 \times 10^{-5})(5 \times 10^{-5}) = 25 \times 10^{-10} = 2.5 \times 10^{-9} \]
- Since \( 2.5 \times 10^{-9} > 1.8 \times 10^{-10} \), precipitation will occur.

**Option (b): \( 10^{-5} M (Ag^{+}) \) and \( 10^{-5} M (Cl^{-}) \)**
- After mixing:
\[ [Ag^{+}] = \frac{10^{-5}}{2} = 5 \times 10^{-6} \, M \]
\[ [Cl^{-}] = \frac{10^{-5}}{2} = 5 \times 10^{-6} \, M \]
- Calculate the ionic product (IP):
\[ IP = (5 \times 10^{-6})(5 \times 10^{-6}) = 25 \times 10^{-12} = 2.5 \times 10^{-11} \]
- Since \( 2.5 \times 10^{-11} < 1.8 \times 10^{-10} \), precipitation will not occur.

**Option (c): \( 10^{-6} M (Ag^{+}) \) and \( 10^{-6} M (Cl^{-}) \)**
- After mixing:
\[ [Ag^{+}] = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, M \]
\[ [Cl^{-}] = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, M \]
- Calculate the ionic product (IP):
\[ IP = (5 \times 10^{-7})(5 \times 10^{-7}) = 25 \times 10^{-14} = 2.5 \times 10^{-13} \]
- Since \( 2.5 \times 10^{-13} < 1.8 \times 10^{-10} \), precipitation will not occur.

**Option (d): \( 10^{-10} M (Ag^{+}) \) and \( 10^{-10} M (Cl^{-}) \)**
- After mixing:
\[ [Ag^{+}] = \frac{10^{-10}}{2} = 5 \times 10^{-11} \, M \]
\[ [Cl^{-}] = \frac{10^{-10}}{2} = 5 \times 10^{-11} \, M \]
- Calculate the ionic product (IP):
\[ IP = (5 \times 10^{-11})(5 \times 10^{-11}) = 25 \times 10^{-22} = 2.5 \times 10^{-21} \]
- Since \( 2.5 \times 10^{-21} < 1.8 \times 10^{-10} \), precipitation will not occur.

### Conclusion
The only option where precipitation occurs is **(a)** \( 10^{-4} M (Ag^{+}) \) and \( 10^{-4} M (Cl^{-}) \).

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