Let's call fluoroacetic acid HFAc. Then
.........HFAc ==> H^+ + FAc^-
E......you do it..0.055..0.055
Ka = (H^+)(FAc^-)/(HFAc)
Substitute the E line into Ka expression and solve for Ka, then convert to pKa.
Note: The acid is 5% dissociated, therefore, every 100 mols dissociates to give 5 of each ion and leaves 95 undissociated molecules. Therefore, M x 0.05% gives you the ion M and 1.11- that gives you the undissociated part.
a 205 mg sample of diprotic acid is dissolved in enough water to make 250 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hyrdoxide (Ksp=1.3*10^-6) is prepared by adding excess calcium hydroxide to
A 205 mg sample of a diprotic acid is dissolved in enough water to make 250.0 ml of solution. The pH of this solution is 2.15. A saturated solution of calcium hydroxide (Ksp= 1.3 x 10-6) is prepared by adding excess calcium
2- A 0.310 M solution of a weak acid, HX, has a pH of 2.53 a. Find the [H+] and the percent ionization of nitrous acid in this solution. b. Write the equilibrium expression and calculate the value of Ka for the weak acid. c.