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Jacqueline Griffith
Answers (1)
pH = pka + log [base CH3COONa]/[acid CH3COOH] 3.7 = 4.7 + log [base]/[acid] alegbra to remove 4.7 from both sides -1.0 = log [base]/[acid] log= 10^x so alegbra to remove 10^x 10^-1 = [base]/[acid] or 0.10 = [base]/[acid] 1 ml for base added to 9 ml of acid
