Question

𝐾𝑎 (𝑁𝐻₄ ⁺) = 10⁻⁹ = 𝑁𝐻₃ [𝐻₃𝑂 ⁺] [𝑁𝐻₄ ⁺] = 𝑥 ∗ 𝑥 0.1 − 𝑥 ≈ 𝑥 2 0.1 ⇒ 𝑥 =

Answer 1

Ans:-

$\to 𝐾𝑎 (𝑁𝐻4 ^+)={NH_3}{[H_3O^+][NH_4^+]}$ for weak electrolyte

$10^{-9}=x^2\times{(0.1-x)}$

We know that $\ \ \ \ 0.1 − 𝑥 ≈ 0.1$

$10^{-9}=x^2\times{0.1}\Rightarrow x=10^{-4}$