Problem: One mole of a weak acid HA was dissolved in 2.0 L of solution. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate Ka for HA.

🤓 Based on our data, we think this question is relevant for Professor Cruz's class at USF.

FREE Expert Solution

To calculate K_a for HA

K_a expression 

$\boxed{{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{A}}^{\mathbf{-}}\mathbf{\right]}}{\mathbf{\left[}\mathbf{HA}\mathbf{\right]}}}$

Step 1: [HA]_initial (molarity)

$$\begin{gathered} \boxed{\mathbf{Molarity}\mathbf{,}\mathbf{ }\mathbf{\left[}\mathbf{HA}\mathbf{\right]}\mathbf{=}\frac{\mathbf{moles}\mathbf{ }\mathbf{HA}}{\mathbf{L}\mathbf{ }\mathbf{solution}}} \\ \mathbf{\left[}\mathbf{HA}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{HA}}{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{L}} \end{gathered}$$

[HA]_initial = 0.5 M