Let HB = benzoic acid.
HB ==> H^+ + B^-
Ka = (H^+)(B^-)/(HB)
Set up an ICE chart.
Convert pH=2.90 to (H^+). Substitute into Ka expression for (H^+) and for (B^-) (they are equal). Solve for (HB) which will be in units of M = moles/L. You know M, you know how many L you want (0.4000), solve for moles. Then moles = grams/molar mass and solve for grams.
What mass of benzoic acid CH6H5COOH, would you dissolve in 400.0 mL of water to produce a solution with a pH= 2.90? Ka= 6.3 x 10^-5
mass in grams is needed.
1 answer