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

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

### Step 2: Define the ionic product (Ksp)
The ionic product (Ksp) for Ni(OH)₂ can be expressed as:
\[ K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2 \]
Given that \( K_{sp} = 1.9 \times 10^{-15} \).

### Step 3: Set up the molar solubility expression
Let the molar solubility of Ni(OH)₂ in 1.0 M NaOH be \( x \). The concentration of Ni²⁺ ions will be \( x \), and the concentration of OH⁻ ions will be affected by the NaOH. Since NaOH is 1.0 M, the total concentration of OH⁻ from NaOH and the dissociation of Ni(OH)₂ will be:
\[ [\text{OH}^-] = 1 + 2x \]
However, since \( x \) is very small compared to 1 M, we can approximate this as:
\[ [\text{OH}^-] \approx 1 \, \text{M} \]

### Step 4: Substitute into the Ksp expression
Substituting the concentrations into the Ksp expression gives:
\[ K_{sp} = [x][1]^2 = x \]
So we have:
\[ 1.9 \times 10^{-15} = x \]

### Step 5: Solve for x
Thus, the molar solubility \( x \) is:
\[ x = 1.9 \times 10^{-15} \, \text{M} \]

### Conclusion
The molar solubility of Ni(OH)₂ in 1.0 M NaOH is \( 1.9 \times 10^{-15} \, \text{M} \).