Question

A benzoic acid/potassium benzoate buffer solution has a pH = 4.25. The concentration of benzoic acid (C6H5COOH, Ka = 6.5 × 10-5) in the solution is 0.54 M. What is the concentration of potassium benzoate (KC6H5COO, Kb = 1.5 × 10-10) in this buffer solution?

Answer 1

Given acidic buffer of (C6H5-COOH + KEH5400) pH of buffer= 4.85 concentration of Acid i.p. [Acid) - 0-sum Ka of And ie. Cons2OOH = 6.5x10-5 pha = -log ka = -log (6.5X10-5) = 4.187 from Henderson - Hasselbalch equation – pH of acidic buffer = pka + 109 [conjugate base] [ Acid] -> 4.95 = 4.187 + 109 [conjugate base] 0.54 5 0.0629 - 109 [conjugare Base] 0.54 taking anti log 1 1.15588 = [conjugate Base] 0.54 3 [conjugare Base J = 0.6241 M concentration of K Co Ms Coo = 0.624m - P.S. - Do not get confused by seeing kn value. It is the same thing represented in different way to create confusion. ký data represents hydrolysis ie. - Аударсу Coo° + но — сон + он°, Khis acting as a base k dr k = kw 10-14 = 1.53810-10 ka 6.5x10.5 we will get same answer even if we follow hydrolysis method instead of using Henderson- Hasselbalch formula.