Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
`AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)]`
Where `K_(sp)` is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as `K_(sp)` are
`QltK_(sp)` Unsaturated solution
`Q=K_(sp)` Saturated solution
`Qgt_(sp)` Supersaturated solution, precipitate will from
Will a precipitate from if 50 `cm^(3)` of 0.01 M `AgNO_(3)` and 50 `cm^(3)` of `2xx10^(-5)` M NaCl are mixed?
`["Given": K_(sp)(AgCl)=10^(-10)M^(2)]`

Options

A. Yes
B. No
C. Ionic product is less than solubility product, hence precipitate will from
D. Data insufficient

Answer

To determine whether a precipitate will form when mixing 50 cm³ of 0.01 M AgNO₃ and 50 cm³ of 2 × 10⁻⁵ M NaCl, we will calculate the reaction quotient (Q) and compare it to the solubility product constant (Ksp) of AgCl, which is given as 10⁻¹⁰ M².

### Step-by-Step Solution:

**Step 1: Calculate the moles of Ag⁺ and Cl⁻ ions.**

1. **Calculate moles of Ag⁺ from AgNO₃:**
- Volume of AgNO₃ = 50 cm³ = 50 × 10⁻³ L = 0.050 L
- Concentration of AgNO₃ = 0.01 M
- Moles of Ag⁺ = Concentration × Volume = 0.01 mol/L × 0.050 L = 5 × 10⁻⁴ moles

2. **Calculate moles of Cl⁻ from NaCl:**
- Volume of NaCl = 50 cm³ = 50 × 10⁻³ L = 0.050 L
- Concentration of NaCl = 2 × 10⁻⁵ M
- Moles of Cl⁻ = Concentration × Volume = 2 × 10⁻⁵ mol/L × 0.050 L = 1 × 10⁻⁶ moles

**Step 2: Calculate the final concentrations of Ag⁺ and Cl⁻ ions in the mixed solution.**

- Total volume of the mixed solution = 50 cm³ + 50 cm³ = 100 cm³ = 0.1 L

1. **Concentration of Ag⁺:**
- [Ag⁺] = Moles of Ag⁺ / Total Volume = (5 × 10⁻⁴ moles) / (0.1 L) = 5 × 10⁻³ M

2. **Concentration of Cl⁻:**
- [Cl⁻] = Moles of Cl⁻ / Total Volume = (1 × 10⁻⁶ moles) / (0.1 L) = 1 × 10⁻⁵ M

**Step 3: Calculate the reaction quotient (Q).**

- Q = [Ag⁺][Cl⁻] = (5 × 10⁻³ M) × (1 × 10⁻⁵ M) = 5 × 10⁻⁸ M²

**Step 4: Compare Q with Ksp.**

- Given Ksp (AgCl) = 10⁻¹⁰ M²
- Since Q = 5 × 10⁻⁸ M² and Ksp = 10⁻¹⁰ M², we find that:
- Q > Ksp

**Conclusion:**

Since Q is greater than Ksp, a precipitate of AgCl will form when the two solutions are mixed.