`K_(sp)` for sodium chloride is `36 mol^(2)//"litre"^(2)`. The solubility of sodium chloride is

Options

A. `(1)/(36)`
B. `(1)/(6)`
C. 6
D. 3600

Explanation

To find the solubility of sodium chloride (NaCl) given its solubility product constant (Ksp), we can follow these steps:

### Step 1: Write the Dissociation Equation
When sodium chloride dissolves in water, it dissociates into sodium ions (Na⁺) and chloride ions (Cl⁻):
\[ \text{NaCl (s)} \rightleftharpoons \text{Na}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of NaCl be represented by \( s \) (in mol/L). This means that at equilibrium:
- The concentration of Na⁺ ions will be \( s \) mol/L.
- The concentration of Cl⁻ ions will also be \( s \) mol/L.

### Step 3: Write the Expression for Ksp
The solubility product constant \( K_{sp} \) is given by the expression:
\[ K_{sp} = [\text{Na}^+][\text{Cl}^-] \]
Substituting the concentrations in terms of solubility \( s \):
\[ K_{sp} = s \cdot s = s^2 \]

### Step 4: Substitute the Given Ksp Value
We are given that \( K_{sp} \) for sodium chloride is 36 mol²/L². Therefore:
\[ s^2 = 36 \]

### Step 5: Solve for Solubility \( s \)
To find \( s \), take the square root of both sides:
\[ s = \sqrt{36} \]
\[ s = 6 \, \text{mol/L} \]

### Conclusion
The solubility of sodium chloride (NaCl) is 6 mol/L.

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