If the solubility product of `MOH` is `1xx10^(-10) mol^(2) dm^(-6)` then `pH` of its aqueous solution will be

Options

A. `12`
B. `9`
C. `6`
D. `3`

Explanation

To find the pH of the aqueous solution of MOH given its solubility product (Ksp), we can follow these steps:

### Step 1: Understand the Dissociation of MOH
When MOH dissolves in water, it dissociates into its ions:
\[ \text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^- \]

### Step 2: Write the Expression for the Solubility Product (Ksp)
The solubility product (Ksp) expression for this dissociation is:
\[ K_{sp} = [\text{M}^+][\text{OH}^-] \]

### Step 3: Substitute the Given Ksp Value
We are given that:
\[ K_{sp} = 1 \times 10^{-10} \, \text{mol}^2 \, \text{dm}^{-6} \]

Assuming the concentration of both ions is equal (let's denote it as \( x \)):
\[ K_{sp} = x \cdot x = x^2 \]
Thus, we have:
\[ x^2 = 1 \times 10^{-10} \]

### Step 4: Solve for x
To find \( x \), we take the square root of both sides:
\[ x = \sqrt{1 \times 10^{-10}} \]
\[ x = 1 \times 10^{-5} \, \text{mol} \, \text{dm}^{-3} \]

This means:
\[ [\text{OH}^-] = 1 \times 10^{-5} \, \text{mol} \, \text{dm}^{-3} \]

### Step 5: Calculate the Concentration of Hydrogen Ions [H+]
We can use the relationship between the concentrations of hydroxide ions and hydrogen ions:
\[ K_w = [\text{H}^+][\text{OH}^-] \]
Where \( K_w = 1 \times 10^{-14} \, \text{mol}^2 \, \text{dm}^{-6} \).

Now, substituting the value of \( [\text{OH}^-] \):
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} \]
\[ [\text{H}^+] = 1 \times 10^{-9} \, \text{mol} \, \text{dm}^{-3} \]

### Step 6: Calculate the pH
Finally, we can calculate the pH using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the value of \( [\text{H}^+] \):
\[ \text{pH} = -\log(1 \times 10^{-9}) \]
\[ \text{pH} = 9 \]

Thus, the pH of the aqueous solution of MOH is **9**.

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