For the following equilibrium
`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-)`
calculate the equilibrium constant, if for the equilibrium,
`NH_(4)^(+)+H_(2)OhArr NH_(4)OH+H^(+)`
the equilibrium constant is `5.5xx10^(-10)`

Options

A. `1.8xx10^(-4)`
B. `1.8xx10^(-5)`
C. `1.8xx10^(-6)`
D. `1.8xx10^(-7)`

Answer

To calculate the equilibrium constant \( K_b \) for the equilibrium reaction:

\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

given that the equilibrium constant \( K_h \) for the reaction:

\[ \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+ \]

is \( 5.5 \times 10^{-10} \).

### Step 1: Write the Equilibrium Constants

1. For the first reaction, we define the equilibrium constant \( K_b \):
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3][\text{H}_2\text{O}]} \]

2. For the second reaction, we define the equilibrium constant \( K_h \):
\[ K_h = \frac{[\text{NH}_4\text{OH}][\text{H}^+]}{[\text{NH}_4^+][\text{H}_2\text{O}]} \]

### Step 2: Relate \( K_b \) and \( K_h \) to \( K_w \)

The relationship between these constants can be expressed as:
\[ K_b \cdot K_h = K_w \]
where \( K_w \) is the ion product of water, which at \( 25^\circ C \) is \( 1.0 \times 10^{-14} \).

### Step 3: Substitute Known Values

We know:
- \( K_h = 5.5 \times 10^{-10} \)
- \( K_w = 1.0 \times 10^{-14} \)

Substituting these values into the equation:
\[ K_b \cdot (5.5 \times 10^{-10}) = 1.0 \times 10^{-14} \]

### Step 4: Solve for \( K_b \)

Now, we can solve for \( K_b \):
\[ K_b = \frac{1.0 \times 10^{-14}}{5.5 \times 10^{-10}} \]

Calculating this gives:
\[ K_b = \frac{1.0}{5.5} \times 10^{-14 + 10} = \frac{1.0}{5.5} \times 10^{-4} \]

Calculating \( \frac{1.0}{5.5} \approx 0.1818 \):
\[ K_b \approx 0.1818 \times 10^{-4} = 1.818 \times 10^{-5} \]

### Step 5: Final Answer

Thus, rounding to two significant figures, we find:
\[ K_b \approx 1.8 \times 10^{-5} \]