`K_(b)` for `NH_(4)OH` is `1.8 xx 10^(-5)`. The `[overset(Theta)OH]` of `0.1 M NH_(4)OH` is

Options

A. `5.0 xx 10^(-2)`
B. `4.20 xx 10^(-3)`
C. `1.34 xx 10^(-3)`
D. `1.8 xx 10^(-6)`

Explanation

To find the hydroxide ion concentration \([OH^-]\) in a 0.1 M solution of \(NH_4OH\) given that \(K_b\) for \(NH_4OH\) is \(1.8 \times 10^{-5}\), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of ammonium hydroxide (\(NH_4OH\)) can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]

### Step 2: Set Up the Initial Concentration
Initially, we have:
- \([NH_4OH] = 0.1 \, M\)
- \([NH_4^+] = 0 \, M\)
- \([OH^-] = 0 \, M\)

### Step 3: Define the Change in Concentration
Let \(x\) be the amount of \(NH_4OH\) that dissociates at equilibrium. Therefore, at equilibrium:
- \([NH_4OH] = 0.1 - x\)
- \([NH_4^+] = x\)
- \([OH^-] = x\)

### Step 4: Write the Expression for \(K_b\)
The expression for the base dissociation constant \(K_b\) is given by:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_4OH]} \]
Substituting the equilibrium concentrations into the expression, we get:
\[ K_b = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

### Step 5: Substitute the Value of \(K_b\)
We know that \(K_b = 1.8 \times 10^{-5}\). Therefore, we can write:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \]

### Step 6: Make an Assumption
Since \(K_b\) is small, we can assume that \(x\) is much smaller than 0.1. Thus, we can simplify \(0.1 - x\) to approximately \(0.1\):
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]
This simplifies to:
\[ x^2 = 1.8 \times 10^{-5} \times 0.1 = 1.8 \times 10^{-6} \]

### Step 7: Solve for \(x\)
Taking the square root of both sides gives:
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 0.00134 \, M \]

### Step 8: Conclusion
Thus, the concentration of hydroxide ions \([OH^-]\) at equilibrium is:
\[ [OH^-] \approx 1.34 \times 10^{-3} \, M \]

### Final Answer
\[ [OH^-] = 1.34 \times 10^{-3} \, M \]