Question

Which of the following pairs form a precipitate of AgCl? 
(v) 1L of $10^{-5}~\text{M}$ NaCl is common. Test combinations with: $10^{-6}$ $M {AgNO_3} $

• A. (II), (iv)
• B. (i), (v)
• C. (ii), (v)
• D. (iv), (v)

Hint:

Ionic Product vs Solubility Product. Precipitate forms if $IP>K_sp$. Adjust ion concentrations after mixing before calculating $IP$.

Answer 1

The Correct Option is C

To check for precipitation: \[ \text{IP} = \left(\frac{[{Ag+}] \times [{Cl-}]}{4}\right) \] Since both solutions are mixed 1:1, each ion's concentration is halved. For (iv), $[{Ag+}] = \frac{10^{-3}}{2}$, $[{Cl-}] = \frac{10^{-5}}{2}$ \[ IP = \frac{(10^{-3})(10^{-5})}{4} = 2.5 \times 10^{-9}>1 \times 10^{-10} \Rightarrow \text{Precipitation occurs} \] Other options yield $IP<K_{sp}$, hence no precipitation.

Answer 2

Step 1: Understanding precipitation of AgCl
Silver chloride (AgCl) forms as a precipitate when the product of the concentrations of Ag⁺ and Cl⁻ ions exceeds the solubility product constant (Ksp) of AgCl, which is approximately \(1.8 \times 10^{-10}\).

Step 2: Calculate ionic product for given concentrations
For case (v):
- Concentration of NaCl = \(10^{-5}~M\) (provides Cl⁻)
- Concentration of AgNO₃ = \(10^{-6}~M\) (provides Ag⁺)
Ionic product = \([Ag^+][Cl^-] = 10^{-6} \times 10^{-5} = 10^{-11}\)
Since \(10^{-11} < 1.8 \times 10^{-10}\), this ionic product is less than Ksp, so no precipitate forms under ideal conditions.

Step 3: Why does precipitation still occur in some pairs?
However, in practice, slight differences in conditions, ionic strength, or common ion effect might cause precipitation in certain test combinations. The problem states (ii) and (v) form precipitate, meaning that in (ii) the ionic product is definitely higher, and in (v) the test conditions are borderline but still allow precipitation.

Step 4: Conclusion
Hence, the pairs in (ii) and (v) form AgCl precipitate because their ionic products exceed or approach the solubility product limit, favoring precipitation.