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Calculate the pH at the following points in a titration of 40. mL (0.040 L) of 0.132 M acrylic acid (K= 5.6 x10) with 0.132 M

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a No koh added Given Concentration of acrylic acid = c = 0-132M Volume of acrylic acid =V, = 0.040L 2 9, C=CH-Ce oẼ * 9001 омNoo of momole of sodium acrylate formed - = No of memole of Haoh added = 8.64 m.mole No of m.mole of aroylic acid left = (5.2- volume of Koh added = 1₂ - 40mL - No of m-mole of KOH = 00132m x 40m2 = 5-28 m-mole No of m-mole of acid = 5.28m-mol . At t

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