a 0.20 mol/l solution of weak base hpo4 has a ph of 9. find the Kb
Actually, you don't need any of this other information since Kb for HPO4- = Kw/Ka for H2PO4^-
To find the Kb (base dissociation constant) of a weak base such as HPO4^2-, you will need to use the pH value and the concentration of the base.
Step 1: Write the balanced chemical equation for the dissociation of the weak base HPO4^2- in water. In this case, HPO4^2- acts as a base and accepts a proton (H+) to form H2PO4^-. The reaction can be written as follows:
HPO4^2- + H2O ⇌ H2PO4^- + OH-
Step 2: Construct the expression for the Kb using the equilibrium concentrations of the species involved:
Kb = [H2PO4^-] * [OH-] / [HPO4^2-]
Step 3: Determine the concentration of OH-. Since the pH of the solution is 9, it means that the concentration of H+ is 10^(-9) mol/L. In water at 25°C, the concentration of OH- and H+ are equal due to the autoionization of water:
[H+] = [OH-] = 10^(-9) mol/L
Step 4: Determine the concentration of H2PO4^-. This can be done by assuming that the initial concentration of HPO4^2- and H2PO4^- is the same because they are in a 1:1 molar ratio. Hence, the concentration of H2PO4^- is also 0.20 mol/L.
Step 5: Plug the values into the Kb expression to calculate the Kb:
Kb = [H2PO4^-] * [OH-] / [HPO4^2-]
Kb = (0.20 mol/L) * (10^(-9) mol/L) / (0.20 mol/L)
Kb = 10^(-9) / (0.20) ≈ 5.0 x 10^(-10)
Therefore, the Kb of the weak base HPO4^2- is approximately 5.0 x 10^(-10).