A monoprotic weak acid in a solution is found to be ionized. What is the value of for this acid? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Given $C = 1.00 \text{ M}$ and $\alpha = 0.01\% = 0.0001 = 10^{-4}$. The dissociation constant is given by $K_a = C\alpha^2$ (since $\alpha \ll 1$). Therefore, $K_a = 1.00 \times (10^{-4})^2 = 10^{-8}$. Multiplying by $10^8$ yields exactly $1$.