Answer to Question #150077 in Inorganic Chemistry for Trigger

Question

You add 0.8 mol of acetic acid (PK=4.7) in 10 ml of water to 0.03 mol of lithium acetate in 100 ml of water .what is the new ph?

Answer 1

pH = pK_a + log([base]/[acid])

C(CH₃COOLi) = 0.03 mol / 0.11 L = 0.27 M

C(CH₃COOH) = 0.8 mol / 0.11 L = 7.27 M

pH = 4.7 + log(0.27 / 7.27) = 3.27