Calculate the `[OH^(-)]` in `0.01M` aqueous solution of `NaOCN(K_(b)` for `OCN^(-)=10^(-10)) :`
(a)`10^(-6)` M
(b)`10^(-7)` M
(c)`10^(-8)` M
(d)None of these

Options

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

Answer

To calculate the hydroxide ion concentration \([OH^-]\) in a \(0.01 \, M\) aqueous solution of \(NaOCN\) (where \(K_b\) for \(OCN^-\) is \(10^{-10}\)), we can follow these steps:

### Step 1: Determine the Hydrolysis Constant \(K_h\)
The hydrolysis constant \(K_h\) for the base \(OCN^-\) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_a} \]
where \(K_w\) is the ion product of water (\(1.0 \times 10^{-14}\) at \(25^\circ C\)) and \(K_a\) is the dissociation constant of the conjugate acid \(HOCN\).

Since we are given \(K_b\) for \(OCN^-\) as \(10^{-10}\), we can find \(K_a\) using the relation:
\[ K_a = \frac{K_w}{K_b} \]
Substituting the values:
\[ K_a = \frac{1.0 \times 10^{-14}}{10^{-10}} = 1.0 \times 10^{-4} \]

### Step 2: Calculate the Hydrolysis Constant \(K_h\)
Now, we can find \(K_h\):
\[ K_h = K_b = 10^{-10} \]

### Step 3: Set Up the Hydrolysis Equation
For the hydrolysis of \(OCN^-\):
\[ OCN^- + H_2O \rightleftharpoons HOCN + OH^- \]
Let \(C\) be the initial concentration of \(NaOCN\), which is \(0.01 \, M\). Let \(x\) be the change in concentration of \(OH^-\) at equilibrium.

### Step 4: Write the Expression for \(K_h\)
The expression for \(K_h\) is:
\[ K_h = \frac{[HOCN][OH^-]}{[OCN^-]} \]
At equilibrium:
- \([HOCN] = x\)
- \([OH^-] = x\)
- \([OCN^-] = C - x \approx C\) (since \(x\) is small compared to \(C\))

Thus, we can write:
\[ K_h = \frac{x^2}{C} \]
Substituting the values:
\[ 10^{-10} = \frac{x^2}{0.01} \]

### Step 5: Solve for \(x\)
Rearranging gives:
\[ x^2 = 10^{-10} \times 0.01 = 10^{-12} \]
Taking the square root:
\[ x = \sqrt{10^{-12}} = 10^{-6} \]

### Step 6: Conclusion
The concentration of hydroxide ions \([OH^-]\) is:
\[ [OH^-] = x = 10^{-6} \, M \]

Thus, the answer is:
**(a) \(10^{-6} \, M\)**.

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