the unknown weak acid having hydronium ion concentration 5.77 x 10⁻¹¹ is to use the formula:pH = pKa + log ([A⁻] / [HA])
Here, HA is the unknown weak acid, and A⁻ is its conjugate base. First, we will find the pH of the solution:Given, hydronium ion concentration = 5.77 x 10⁻¹¹pH = -log[H₃O⁺] = -log(5.77 x 10⁻¹¹) = 10.239Now, we will use the formula to find the pKa:
pH = pKa + log ([A⁻] / [HA])10.239 = pKa + log ([A⁻] / [HA])We know that the solution is a weak acid, so the conjugate base concentration is negligible. Hence,[A⁻] ≈ 0Substituting the values, we get:10.239 = pKa + log (0 / [HA])10.239 = pKa - ∞10.239 + ∞ = pKapKa = ∞This means that the pKa of the unknown weak acid is very high, so it is a very weak acid.