Let's call potassium sorbate KSor just to keep the typing to a minimum.
mols HSor = grams/molar mass = 2.46/112 = 0.022
M HSor = moles/L = 0.022/0.250 = 0.0878 M
........................Sor^- + H2O ==> HSor + OH^-
I......................0.0878......................0...........0
C........................-x...........................x...........x
E..................0.0878-x.......................x...........x
Kb for Sorbate ion = Kw/Ka for sorbic acid = (x)(x)/(0.0878-x)
You know Kw and Ka. Plug those in and solve for x = (OH^-) and convert to pH. Post your work if you get stuck.
Sorbic acid, HC6H7O2, is a weak monoprotic acid, with Ka = 1.7 × 10-5. What is the pH of a
solution containing 2.46 g of potassium sorbate in 250 cm3 of solution?
1 answer
2 answers