Calculate the `[OH^(-)]` in `0.01M` aqueous solution of `NaOCN(K_(b)` for `OCN^(-)=10^(-10)) :`

Options

A. `10^(-6)M`
B. `10^(-7)M`
C. `10^(-8)M`
D. none of these

Explanation

To calculate the concentration of hydroxide ions \([OH^-]\) in a \(0.01 M\) aqueous solution of sodium cyanate \((NaOCN)\), we will follow these steps:

### Step 1: Understanding the Hydrolysis of Sodium Cyanate
Sodium cyanate \((NaOCN)\) is a salt that dissociates in water to give sodium ions \((Na^+)\) and cyanate ions \((OCN^-)\). The cyanate ion is the conjugate base of the weak acid, cyanic acid \((HOCN)\). The hydrolysis of the cyanate ion can be represented as:
\[ OCN^- + H_2O \rightleftharpoons HOCN + OH^- \]
This reaction shows that the cyanate ion will produce hydroxide ions in solution.

### Step 2: Setting Up the Hydrolysis Constant
The hydrolysis constant \((K_h)\) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_b} \]
where \(K_w\) is the ion product of water \((1.0 \times 10^{-14})\) and \(K_b\) is the base dissociation constant for the cyanate ion, given as \(10^{-10}\).

Calculating \(K_h\):
\[ K_h = \frac{1.0 \times 10^{-14}}{10^{-10}} = 1.0 \times 10^{-4} \]

### Step 3: Applying the Hydrolysis Constant
For the hydrolysis reaction, we can express the concentration of hydroxide ions \([OH^-]\) in terms of the initial concentration of the cyanate ion \((C)\) and the hydrolysis constant \((K_h)\):
\[ K_h = \frac{[HOCN][OH^-]}{[OCN^-]} \]
Assuming \(x\) is the concentration of \([OH^-]\) produced, we can express the equilibrium concentrations as follows:
- \([HOCN] = x\)
- \([OH^-] = x\)
- \([OCN^-] = C - x \approx C\) (since \(x\) will be small compared to \(C\))

Substituting these into the expression for \(K_h\):
\[ K_h = \frac{x \cdot x}{C} = \frac{x^2}{C} \]
Given \(C = 0.01 M\):
\[ 1.0 \times 10^{-4} = \frac{x^2}{0.01} \]

### Step 4: Solving for \([OH^-]\)
Rearranging the equation:
\[ x^2 = 1.0 \times 10^{-4} \cdot 0.01 \]
\[ x^2 = 1.0 \times 10^{-6} \]
Taking the square root:
\[ x = \sqrt{1.0 \times 10^{-6}} = 1.0 \times 10^{-3} \]

Thus, the concentration of hydroxide ions \([OH^-]\) is:
\[ [OH^-] = 1.0 \times 10^{-3} M \]

### Final Answer
The concentration of hydroxide ions in the \(0.01 M\) aqueous solution of sodium cyanate is:
\[ \boxed{1.0 \times 10^{-3} M} \]