If the ionic product of Ni `(OH)_2` is `1.9xx10^(-15)`, the molar solubility of `Ni (OH)_2` in 1.0 M NaOH is

Options

A. `1.9 xx10^(-18)M`
B. `1.9 xx10^(-13)M`
C. `1.9 xx10^(-15)M`
D. `1.9 xx10^(-14)M`

Answer

To find the molar solubility of Ni(OH)₂ in 1.0 M NaOH, we can follow these steps:

### Step 1: Write the dissociation equation for Ni(OH)₂
Nickel hydroxide dissociates in water as follows:
\[ \text{Ni(OH)}_2 (s) \rightleftharpoons \text{Ni}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define the solubility
Let the molar solubility of Ni(OH)₂ be \( S \). Therefore, at equilibrium:
- The concentration of \( \text{Ni}^{2+} \) ions will be \( S \).
- The concentration of \( \text{OH}^- \) ions will be \( 2S \) (since 2 moles of OH⁻ are produced for every mole of Ni(OH)₂ that dissolves).

### Step 3: Consider the contribution of NaOH
Since we are in a 1.0 M NaOH solution, the concentration of OH⁻ ions from NaOH is 1.0 M. Therefore, the total concentration of OH⁻ ions in the solution will be:
\[ \text{Total } [\text{OH}^-] = 2S + 1 \]

### Step 4: Write the expression for the solubility product (Ksp)
The solubility product \( K_{sp} \) for Ni(OH)₂ can be expressed as:
\[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (2S + 1)^2 \]

### Step 5: Substitute the given Ksp value
We know that \( K_{sp} \) for Ni(OH)₂ is given as \( 1.9 \times 10^{-15} \):
\[ 1.9 \times 10^{-15} = S \cdot (2S + 1)^2 \]

### Step 6: Make an approximation
Since \( K_{sp} \) is very small, we can assume that \( S \) is much smaller than 1. Therefore, we can neglect \( S \) in the expression \( (2S + 1) \):
\[ 1.9 \times 10^{-15} \approx S \cdot (1)^2 \]
Thus,
\[ S \approx 1.9 \times 10^{-15} \]

### Step 7: Conclusion
The molar solubility of Ni(OH)₂ in 1.0 M NaOH is:
\[ S \approx 1.9 \times 10^{-15} \, \text{M} \]