This problem involves the concept of solubility equilibria and the common ion effect. We have two sparingly soluble salts, AgCl and CuCl, sharing a common ion, Cl⁻. When CuCl is added to a saturated solution of AgCl, the chloride ions from CuCl will suppress the solubility of AgCl due to the common ion effect. We need to find the new concentration of Ag⁺ ions.

Step 1: Understand the Initial System (Saturated AgCl Solution)

For AgCl dissolving in water: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

The solubility product constant is given by: $K_{\mathrm{sp}}\left(\text{AgCl}\right)=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=1.6\times {10}^{-10}$

In a pure saturated solution of AgCl, let the solubility be 's' mol/L. Then, [Ag⁺] = s and [Cl⁻] = s. $K_{\mathrm{sp}}=s\times s=s^2=1.6\times {10}^{-10}$
Therefore, initial $\left[{\mathrm{Cl}}^{-}\right]=s=\sqrt{1.6\times {10}^{-10}}=1.2649\times {10}^{-5}\text{M}$. This is very small.

Step 2: Analyze the Addition of CuCl

We are adding 0.1 mol of CuCl to 1 L of this saturated AgCl solution. The Ksp of CuCl is much larger (1.0 × 10⁻⁶) than that of AgCl. This means CuCl is significantly more soluble.

The dissolution reaction for CuCl is: CuCl(s) ⇌ Cu⁺(aq) + Cl⁻(aq)

When we add 0.1 mol of solid CuCl to the solution, a large amount of it will dissolve because its Ksp is relatively high. The chloride ions from this dissolution will be the major source of Cl⁻ in the solution, overwhelming the tiny amount coming from AgCl. The common ion effect will strongly suppress the dissolution of AgCl.

Key Assumption: The concentration of Cl⁻ in the solution is now effectively determined by the dissolved CuCl, as it provides a much higher concentration. We can assume that the [Cl⁻] from dissolved CuCl is approximately 0.1 M. Let's verify this assumption.

Let the solubility of CuCl in this mixture be 'y'. Then, it would contribute [Cu⁺] = y and [Cl⁻] = y. However, we are adding a large amount (0.1 mol), so 'y' will be significant. But we must also consider that the solution already has a tiny amount of Cl⁻ from AgCl. The total [Cl⁻] would be y (from CuCl) + a very small amount from AgCl.

But since Ksp(CuCl) = 1.0 × 10⁻⁶ = [Cu⁺][Cl⁻], if [Cl⁻] is around 0.1 M, then [Cu⁺] would be (1.0 × 10⁻⁶)/0.1 = 1.0 × 10⁻⁵ M. This means only 1.0 × 10⁻⁵ mol of CuCl has dissolved to maintain equilibrium, which is negligible compared to the 0.1 mol we added. Therefore, the solution is effectively saturated with CuCl, and the [Cl⁻] is dominated by the chloride from the dissolved portion of the added 0.1 mol of CuCl.

For CuCl: $K_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=1.0\times {10}^{-6}$

Since the CuCl(s) is present in excess (0.1 mol >> 10⁻⁵ mol), the solution will be saturated with respect to CuCl. In such a saturated solution, the product [Cu⁺][Cl⁻] must equal 1.0 × 10⁻⁶.

Step 3: Find the New Chloride Ion Concentration

The total [Cl⁻] in the solution comes from two sources: dissolved CuCl and dissolved AgCl. Let [Cl⁻] from CuCl be 'a' and from AgCl be 'b'. So, [Cl⁻]_total = a + b.

From the dissolution equations: $\left[{\mathrm{Cu}}^{+}\right]=a$ and $\left[{\mathrm{Ag}}^{+}\right]=b$.

We can write the Ksp expressions: $K_{\mathrm{sp}}\left(\text{CuCl}\right)=a\times (a+b)=1.0\times {10}^{-6}$
$K_{\mathrm{sp}}\left(\text{AgCl}\right)=b\times (a+b)=1.6\times {10}^{-10}$

Notice that both equations share the term (a + b), which is the total [Cl⁻]. Since Ksp(AgCl) is very small, 'b' (the solubility of AgCl, which is [Ag⁺]) will be very small. Therefore, (a + b) ≈ a. This is the common ion effect: the chloride from CuCl dominates.

So, we can approximate: $K_{\mathrm{sp}}\left(\text{CuCl}\right)\approx a\times a=a^2=1.0\times {10}^{-6}$
Thus, $a\approx \sqrt{1.0\times {10}^{-6}}={10}^{-3}\text{M}$.

Therefore, the total chloride concentration is [Cl⁻] ≈ a = 10⁻³ M.

Step 4: Find the New Silver Ion Concentration

Now we use the Ksp for AgCl with this new [Cl⁻] value. $K_{\mathrm{sp}}\left(\text{AgCl}\right)=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=1.6\times {10}^{-10}$
$\left[{\mathrm{Ag}}^{+}\right]\times \left({10}^{-3}\right)=1.6\times {10}^{-10}$
$\left[{\mathrm{Ag}}^{+}\right]=\frac{1.6\times {10}^{-10}}{{10}^{-3}}=1.6\times {10}^{-7}\text{M}$

The problem states that this resultant concentration is 1.6 × 10⁻ˣ. Comparing, we find 1.6 × 10⁻ˣ = 1.6 × 10⁻⁷. Therefore, x = 7.

Final Answer: The value of "x" is 7.

Related Topics & Formulae

Solubility Product Constant (Ksp): For a general salt A_aB_b(s) ⇌ aAⁿ⁺(aq) + bB^m-(aq), the solubility product is Ksp = [Aⁿ⁺]^a[B^m-]^b. It is the equilibrium constant for the dissolution of a sparingly soluble salt.

Common Ion Effect: The solubility of a salt is decreased when another salt, which has an ion in common with it, is added to the solution. This is due to Le Chatelier's principle; adding the product ion shifts the equilibrium towards the solid reactant.