Answer to Question #150077 in Inorganic Chemistry for Trigger

Question #150077
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?
1
Expert's answer
2020-12-11T01:45:52-0500

pH = pKa + 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


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