The base ionization constant, K_b, of ammonia in water is $$1.8×10^{-5}$$. ​The value of acid ionization constant, K_a, of the conjugate acid is closest to

​

Correct option is A

For ammonia (NH₃) and its conjugate acid ammonium ion (NH₄⁺), we have the following equilibrium reactions in water:

$$\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$$​

The base ionization constant, K_b, is given as 1.8×10^-5. It represents the equilibrium constant for the reaction of ammonia with water to form ammonium ions and hydroxide ions. The acid ionization constant, K_a, for the conjugate acid ammonium ion (NH₄⁺), is related to K_b through the following expression:

$$K_a×K_b=K_w$$​

where K_w is the ionic product constant for water at a given temperature, which is approximately 1.0×10^-14 at room temperature. The value of K_a can be calculated as follows:

$$K_a×K_b=K_w$$

or, $$K_a×(1.8×10{-5})=1.0×10^{-14}$$

or, $$K_a=(1.0×10^{-14})/(1.8×10^{-5})$$​ 

or, $$K_a \approx 5.6×10^{-10}$$​