Map A macmilan learning Sapling Learning A 100.0 mL solution containing 0.755 g

Question

Map A macmilan learning Sapling Learning A 100.0 mL solution containing 0.755 g of maleic acid (MW 116.072 g/mol) is titrated with 0.217 M KOH. Calculate the pH of the solution after the addition of 60.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. Number 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 H2M, HM and M which represent the fully protonated, intermediate and fully deprotonated forms, respectively. Number M Number M Number M

Explanation / Answer

mmol of acid = mass/MW= 0.755/116.072 = 0.0065 mol of acid = 6.5 mmol

mmol of base = MV = 0.217*60 = 13 mmol

so..

mmol of H+ =13

mmol of OH- = 13

so this is neutralization

so...

A-2 + H2O <--< HA + OH-

Kb2 = Kw/Ka2 = (10^-14)/(10^-6.27) = 1.86208*10^-8

Kb2 =[HA][OH-][A-2]

total V = 100+60 = 160

[A-2] = mmol/V = 6.5/160 = 0.040625

Kb2 =[HA][OH-][A-2]

1.86208*10^-8 = x*x/(0.040625-x)

x = [OH-] = 2.726*10^-5

so..

[M-2] = 0.040625 - 2.726*10^-5 = 0.04059774 M

for

[HM-] = 2.726*10^-5

second

Kb1 = [H2A][OH-][HA-]

(10^-14)/(10^-1.92) = 8.31763*10^-13

8.31763*10^-13 = y*y/(2.726*10^-5-y)

y = 4.76*10^-9

so

[H2M] = 4.76*10^-9

[HM-] = 2.726*10^-5-4.76*10^-9 = 2.725*10^-5

pOH = -log(2.726*10^-5) = 4.564

pH = 14-pOH = 14-4.564 = 9.436

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