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A 100.0 mL solution containing 0.852 g of maleic acid (MW = 116.072 g/mol) is titrated...

Question

A 100.0 mL solution containing 0.852 g of maleic acid (MW = 116.072 g/mol) is titrated...

A 100.0 mL solution containing 0.852 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.277 M KOH. Calculate the pH of

A 100.0 mL solution containing 0.852 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.277 M KOH. Calculate the pH of the solution after the addition of 53.0 mL of the KOH solution. Maleic acid has pK, values of 1.92 and 6.27. pH = -.57 At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as H, M, HMⓇ, and M2-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M?-1 = 3.72 [HM") = 3.75 [H,M] = 7.32

Answers

Ch mules malec as TSres vaume (n) malelc acd,M anAa moles 8529 t16-0329/m) 2O-0033Ymules Moles tCon Cadded O-233 m53-0 m13 O0CHM Cm)t C +Cm Cdvde numvhy deromnokw Am by MA Carm CM 2 CAM t Cm CHA CA CMm CH KeA Kar ICa CH CHm Cm)t CHt Cm 2 Kai CHTS CHmCAZA 3-9&1 XO-0480 2O03836 m 4-98100 6-038m CHm 2 X O-0480m O Oo963 m 4-98104 Kaz Cut Cm2) 1-98101 122 K1O 2. XO-0480Mm 4-481


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