What molar concentration of `NH_3` provides a (OH) of `1.5 xx 10^(-3)? (K_(b)=1.8 xx 10^(-5))`.

Options

A. 0.125
B. `(0.125+1.5 xx 10^(-3))`
C. `(0.125-1.5 xx 10^(-3))`
D. `(1.5 xx 10^(-3))`

Answer

To find the molar concentration of ammonia (NH₃) that provides a hydroxide ion concentration of \( [OH^-] = 1.5 \times 10^{-3} \) M, we can use the base dissociation constant \( K_b \) for ammonia, which is given as \( K_b = 1.8 \times 10^{-5} \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the dissociation of NH₃:**
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
The equilibrium expression for this reaction is:
\[ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \]

2. **Identify the concentrations at equilibrium:**
Let \( [NH_3] \) be the concentration of ammonia at equilibrium, and since \( [OH^-] \) is given as \( 1.5 \times 10^{-3} \) M, we can assume that \( [NH_4^+] = [OH^-] = 1.5 \times 10^{-3} \) M at equilibrium.

3. **Substitute into the \( K_b \) expression:**
\[ K_b = \frac{(1.5 \times 10^{-3})(1.5 \times 10^{-3})}{[NH_3]} \]
\[ 1.8 \times 10^{-5} = \frac{(1.5 \times 10^{-3})^2}{[NH_3]} \]

4. **Calculate \( (1.5 \times 10^{-3})^2 \):**
\[ (1.5 \times 10^{-3})^2 = 2.25 \times 10^{-6} \]

5. **Substitute this value back into the equation:**
\[ 1.8 \times 10^{-5} = \frac{2.25 \times 10^{-6}}{[NH_3]} \]

6. **Rearranging to solve for \( [NH_3] \):**
\[ [NH_3] = \frac{2.25 \times 10^{-6}}{1.8 \times 10^{-5}} \]

7. **Perform the division:**
\[ [NH_3] = \frac{2.25}{1.8} \times 10^{-6 + 5} = 1.25 \times 10^{-1} = 0.125 \, M \]

### Final Answer:
The molar concentration of \( NH_3 \) that provides a hydroxide ion concentration of \( 1.5 \times 10^{-3} \) M is \( 0.125 \, M \).