To solve these problems, we need to use the relationship between the acid dissociation constant ($K_a$) and the base dissociation constant ($K_b$) for conjugate acid-base pairs. This relationship is given by:

$K_a \times K_b = K_w$

where $K_w$ is the ionization constant of water at 25°C, which is $1.0 \times 10^{-14}$.

Problem 1: $K_b$ for $\text{BrO}^-$

Given:

• $K_a$ for $\text{HBrO}$ = $2.8 \times 10^{-9}$

We can find $K_b$ for $\text{BrO}^-$ by rearranging the equation:

$K_b = \frac{K_w}{K_a}$

Substituting the given values:

$K_b = \frac{1.0 \times 10^{-14}}{2.8 \times 10^{-9}} = 3.57 \times 10^{-6}$

So, $K_b$ for $\text{BrO}^-$ is $3.57 \times 10^{-6}$.

Problem 2: $K_a$ for $\text{NH}_4^+$

Given:

• $K_b$ for $\text{NH}_3$ = $1.8 \times 10^{-5}$

Similarly, we can find $K_a$ for $\text{NH}_4^+$ using:

$K_a = \frac{K_w}{K_b}$

Substituting the given values:

$K_a = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

Thus, $K_a$ for $\text{NH}_4^+$ is $5.56 \times 10^{-10}$.

Would you like a deeper explanation of any part of these calculations?

Related Questions

1. What is the significance of $K_a$ and $K_b$ in acid-base chemistry?
2. How does temperature affect the value of $K_w$?
3. Can $K_a$ or $K_b$ alone determine if a solution is acidic or basic?
4. What are conjugate acid-base pairs, and how are they determined?
5. Why is $K_w$ always constant at a specific temperature?

Tip: Always check units and significant figures when dealing with equilibrium constants in calculations.