Question

when 1.00 g of stearic acid ( a saturated fatty acid) is dissolved in water, then diluted to exctly 10.00 ml, the PH of the resulting solution is 2.73.the fornula weight of steric acid is 284.5. calculate ka and pka for this acid

Answer 1

pH = 2.73 = -log(H^+). Solve for (H^+) and I get something like 1.86E-3 but you need to confirm that. Let HA = fatty acid.
.........HA --> H^+ + A^-
.........x.......0....0
equil....x......1.86E-3..1.86E-3
Ka = (H^+)(A^-)/(HA)
Substitute into Ka expression and solve for x.