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saplnglearning.com/ibiscms/modibiswiew phprid 3413330 options No results Sapling Learning A 100.0 mL solution containing 0.9819gof maleic acid (MW 116.072 g/mol) is titrated with 0.3192 M KOH. Calculate the pH of the solution atter the addition of 53.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27 pH 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 M2 which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. M M HM H,M O Previous 8Give up s view solution check Answer Next Ext about us careers privacy policy terms of use contact us help Gra O H OT

Explanation / Answer

The reactions occur as follows:

H2M + OH- --------> HM- + H2O Ka1

HM- +  OH- ----------> M2- + H2O Ka2

moles of maleic acid = 0.9819 g / 116.072 g/mol = 8.46 X 10-3 moles.

moles of KOH = 0.3192 mol/L X 0.053 L = 0.017 moles.

We can tell from the reaction that 2 equivalents of base will be needed to completely neutralize the acid. 1 equivalent of the base will neutralize half the acid at the first equivalent point and another equivalent of base will neutralize the remaining half of the acid.

moles of KOH needed to reach the first equivalence point = 8.46 X 10-3 moles X 1 = 8.46 X 10-3 moles.

volume of KOH needed to reach the first equivalence point = 8.46 X 10-3 moles / 0.3192 M = 0.027 L or 27 ml

moles of KOH needed to reach the second equivalence point = 8.46 X 10-3 moles X 2 = 0.017 moles.

volume of KOH needed to reach the second equivalence point = 0.017 moles / 0.3192 M = 0.053 L or 53 ml.

Total volume at second equivalence point = 0.053 L KOH + 0.1 L maleic acid = 0.153 L

molarity of KOH at second equivalence point = 0.017 moles / 0.153 L = 0.111 M

Now that we know all the volumes and molarities at the two equivalence points of the titration let's move on to calculate the pH after addition of 53 ml of KOH.

We have been given pKa2 = 6.27. Thus, pKa2 = -log Ka2. So, Ka2 = 5.37 X 10-7

We know that Kb1 = Kw / Ka2

Kb1 = 1 X 10-14 / 5.37 X 10-7

Kb1 = 1.862 X 10-8

[OH-] = (Kb1 X molarity of KOH at second equivalence point)1/2

[OH-] = ( 1.862 X 10-8 X 0.111 M)1/2

[OH-] = 4.546 X 10-5 M

pOH = -log [OH-]

pOH = 4.342

pH = 14 - pOH = 9.658

Thus, pH after addition of 53 ml KOH solution is 9.658.

We have been specifically asked to calculate concnetrations of species AFTER the second equivalence point.

[H2M] = number of moles remaining / total volume at second equivalence point

[H2M] = 0 / 0.153 L = 0 M

[HM-] = number of moles remaining / total volume at second equivalence point

[HM-] = 0 / 0.153 L =0 M

[M2-] : Ka2 = [M2-] / [HM-][OH-]

5.37 X 10-7 = x / (8.589 X 10-7) (0.017)

x = 7.841 X 10-15 moles

[M2-] = number of moles of [M2-] after completion of reaction 2 + excess moles of KOH still remaining in the solution after completion of reaction 2

[M2-] = 7.841 X 10-15 moles + 8.459 X 10-3 moles

[M2-] = 8.459 X 10-3 moles

[M2-] =  8.459 X 10-3 moles / total volume after reaching second equivalence point

[M2-] =  8.459 X 10-3 moles / 0.153 L = 0.055 M

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