An aqueous solution of 0.21 M ammonia (NH3) has a pOH of 2.72. Determine the base-dissociation constant (Kb) of ammonia.


To determine the base dissociation constant (\(K_b\)) of ammonia (\(NH_3\)) from the given information, we can follow these steps:

### Step 1: Use the pOH to find the hydroxide ion concentration \([\text{OH}^-]\)

We are given that the pOH is 2.72. The concentration of hydroxide ions \([\text{OH}^-]\) can be found using the formula:

\[
[\text{OH}^-] = 10^{-\text{pOH}}
\]

Substitute the given pOH value:

\[
[\text{OH}^-] = 10^{-2.72}
\]

Now, let's calculate that value.

\[
[\text{OH}^-] = 1.905 \times 10^{-3} \, \text{M}
\]

### Step 2: Set up the base dissociation equation for ammonia

The base dissociation of ammonia in water is represented as:

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

Let \(x\) represent the concentration of \([\text{OH}^-]\), which is also the concentration of \([NH_4^+]\) due to the stoichiometry of the reaction. This means that:

\[
x = [\text{OH}^-] = 1.905 \times 10^{-3} \, \text{M}
\]

### Step 3: Use the ICE table to set up the expression for \(K_b\)

| Species      | Initial (M)      | Change (M)       | Equilibrium (M)  |
|--------------|------------------|------------------|------------------|
| \(NH_3\)     | 0.21             | \(-x\)           | \(0.21 - x\)     |
| \(NH_4^+\)   | 0                | \(+x\)           | \(x\)            |
| \(OH^-\)     | 0                | \(+x\)           | \(x\)            |

Since \(x = [\text{OH}^-]\), we have:

\[
x = 1.905 \times 10^{-3} \, \text{M}
\]

The equilibrium concentration of \(NH_3\) is:

\[
[ NH_3 ] = 0.21 - 1.905 \times 10^{-3} \approx 0.2081 \, \text{M}
\]

### Step 4: Write the expression for \(K_b\)

The expression for the base dissociation constant \(K_b\) is:

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

Substituting the values:

\[
K_b = \frac{(1.905 \times 10^{-3})(1.905 \times 10^{-3})}{0.2081}
\]

Now, calculate the value of \(K_b\):

\[
K_b = \frac{3.630 \times 10^{-6}}{0.2081} = 1.744 \times 10^{-5}
\]

### Final Answer:

The base dissociation constant \(K_b\) for ammonia is:

\[
K_b = 1.74 \times 10^{-5}
\]

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How is the relationship between pOH and pH used to find the dissociation constants?
2. What role does ammonia's structure play in its basicity in water?
3. How can the \(K_b\) of a base be related to the \(K_a\) of its conjugate acid?
4. How does changing the concentration of ammonia affect the pOH of the solution?
5. What are the units of the base dissociation constant \(K_b\)?

**Tip:** The relationship between pH and pOH can always be used to find unknowns, as they are connected by \( \text{pH} + \text{pOH} = 14 \).

An aqueous solution of 0.21 M ammonia (NH3) has a pOH of 2.72. Determine the base-dissociation constant (Kb) of ammonia.

Learn how to determine the base dissociation constant (Kb) for ammonia (NH3) with a pOH of 2.72 and 0.21 M concentration. This step-by-step solution covers key concepts such as equilibrium, logarithms, and acid-base chemistry. Suitable for high school students in Grades 10-12.

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