Ionization Constants Relationship in Water

In aqueous solutions, there is a fundamental relationship between the acid ionization constant (K_a) of an acid and the base ionization constant (K_b) of its conjugate base. This relationship is governed by the ion product of water (K_w).

The ion product of water, K_w, is the equilibrium constant for the autoionization of water:

$\text{H}_2\text{O(l)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{OH}^-\text{(aq)}$

At 25°C, the value of K_w is approximately $1.0 \times 10^{-14}$.

For a conjugate acid-base pair, the product of their ionization constants in water is equal to K_w:

$\text{K}_\text{a} \times \text{K}_\text{b} = \text{K}_\text{w}$

Ammonia and Its Conjugate Acid

The question provides the base ionization constant, K_b, for ammonia ($\text{NH}_3$) in water. Ammonia acts as a base by accepting a proton from water:

$\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$

In this reaction, $\text{NH}_3$ is the base, and $\text{NH}_4^+$ (ammonium ion) is its conjugate acid.

We are given the K_b value for ammonia:

$\text{K}_\text{b} (\text{NH}_3) = 1.8 \times 10^{-5}$

We need to find the acid ionization constant, K_a, of its conjugate acid, $\text{NH}_4^+$. The equilibrium reaction for $\text{NH}_4^+$ acting as an acid in water is:

$\text{NH}_4^+\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}$

or simply showing the proton donation:

$\text{NH}_4^+\text{(aq)} \rightleftharpoons \text{NH}_3\text{(aq)} + \text{H}^+\text{(aq)}$

Calculating the Ionization Constant

Using the relationship $\text{K}_\text{a} \times \text{K}_\text{b} = \text{K}_\text{w}$, we can solve for K_a:

$\text{K}_\text{a} (\text{NH}_4^+) = \frac{\text{K}_\text{w}}{\text{K}_\text{b} (\text{NH}_3)}$

Assuming the temperature is 25°C, K_w = $1.0 \times 10^{-14}$.

Substitute the given values:

$\text{K}_\text{a} (\text{NH}_4^+) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}$

Now, perform the calculation:

$\text{K}_\text{a} = \frac{1.0}{1.8} \times \frac{10^{-14}}{10^{-5}}$

$\text{K}_\text{a} \approx 0.555... \times 10^{(-14) - (-5)}$

$\text{K}_\text{a} \approx 0.555... \times 10^{-14 + 5}$

$\text{K}_\text{a} \approx 0.555... \times 10^{-9}$

To express this in standard scientific notation (a value between 1 and 10 multiplied by a power of 10), we move the decimal one place to the right:

$\text{K}_\text{a} \approx 5.55... \times 10^{-10}$

Rounding this value to a reasonable number of significant figures (usually two, based on the given Kb value), we get $5.6 \times 10^{-10}$.

Final Ionization Constant Value

The calculated value for the acid ionization constant, K_a, of the conjugate acid ($\text{NH}_4^+$) is approximately $5.6 \times 10^{-10}$. Comparing this to the options provided, the closest value is $5.6 \times 10^{-10}$.