What is the approximate `OH^(-)` ion concentration of a `0.150M NH_(3)` solution? `(K_(b)=1.75xx10^(-5))`

Options

A. `2.62xx10^(-6)`
B. `4.6xx10^(-6)`
C. `1.62xx10^(-3)`
D. `3.6xx10^(-3)`

Answer

To find the approximate concentration of `OH^(-)` ions in a `0.150 M NH3` solution, we can follow these steps:

### Step 1: Write the equilibrium reaction
Ammonia (`NH3`) acts as a weak base and reacts with water (`H2O`) to produce ammonium ions (`NH4^+`) and hydroxide ions (`OH^(-)`):

\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]

### Step 2: Set up the initial concentrations
Initially, we have:
- \([NH_3] = 0.150 \, M\)
- \([NH_4^+] = 0\)
- \([OH^-] = 0\)

### Step 3: Define the change in concentration at equilibrium
Let `x` be the change in concentration of `OH^(-)` ions at equilibrium. Therefore, at equilibrium, we have:
- \([NH_3] = 0.150 - x\)
- \([NH_4^+] = x\)
- \([OH^-] = x\)

### Step 4: Write the expression for the base dissociation constant \( K_b \)
The expression for \( K_b \) for the reaction is given by:

\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \]

Substituting the equilibrium concentrations into the expression:

\[ K_b = \frac{x \cdot x}{0.150 - x} = \frac{x^2}{0.150 - x} \]

### Step 5: Substitute the value of \( K_b \)
Given \( K_b = 1.75 \times 10^{-5} \):

\[ 1.75 \times 10^{-5} = \frac{x^2}{0.150 - x} \]

### Step 6: Make an assumption
Since \( K_b \) is small, we can assume that \( x \) is very small compared to `0.150`, allowing us to simplify the equation:

\[ 1.75 \times 10^{-5} \approx \frac{x^2}{0.150} \]

### Step 7: Solve for \( x^2 \)
Rearranging gives:

\[ x^2 = 1.75 \times 10^{-5} \times 0.150 \]

Calculating this:

\[ x^2 = 2.625 \times 10^{-6} \]

### Step 8: Calculate \( x \)
Taking the square root to find \( x \):

\[ x = \sqrt{2.625 \times 10^{-6}} \approx 1.62 \times 10^{-3} \, M \]

### Conclusion
The approximate concentration of `OH^(-)` ions in the `0.150 M NH3` solution is:

\[ [OH^-] \approx 1.62 \times 10^{-3} \, M \]

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