Question

Calculate the pH of a solution that contains 2.7 M HF and 2.7 M HOC6H5. Also, calculate the concentration of OC6H5- in this solution at equilibrium. Ka(HF) = 7.2×10-4; Ka(HOC6H5) = 1.6×10-10. pH = [OC6H5-] = M

Answer 1

To calculate the pH of a solution containing 2.7 M HF and 2.7 M HOC6H5, we can use the Ka values of the acids to set up an ICE table and calculate the equilibrium concentrations of the ions. The concentration of H3O+ ions will give us the pH, and we can calculate the concentration of OC6H5- ions at equilibrium.

Step-by-step explanation:

To calculate the pH of a solution containing 2.7 M HF and 2.7 M HOC6H5, we need to consider the ionization of both compounds. HF is a weak acid and HOC6H5 is an even weaker acid. We will use the Ka values of HF and HOC6H5 to calculate the equilibrium concentrations of the ions at equilibrium. The equilibrium concentrations of HF and HOC6H5 can be represented as [HF] = 2.7 - x and [HOC6H5] = 2.7 - x, respectively, where x is the concentration of the ions at equilibrium.

Using the Ka values, we can set up an ICE table to calculate the equilibrium concentrations of the ions. For HF, the equation will be HF + H2O ⟶ H3O+ + F-. For HOC6H5, the equation will be HOC6H5 + H2O ⟶ H3O+ + OC6H5-.

From the ICE table, we can calculate the concentration of H3O+ ions, which will give us the pH of the solution. Finally, we can calculate the concentration of OC6H5- ions at equilibrium.