A 100.0 mL 100.0 mL solution containing 0.960 g 0.960 g of maleic acid (MW = 116.072 g/mol) (MW=116.072 g/mol) is titrated with 0.295 M KOH. 0.295 M KOH. Calculate the pH of the solution after the addition of 56.0 mL 56.0 mL of the KOH KOH solution. Maleic acid has p K a pKa values of 1.92 and 6.27. pH =

pH = 9.5

Explanation:

The strategy here is to calculate how many moles of maleic acid will react after titration with KOH.

We are given the MWs so we can do the stoichiometric calculation for the reaction:

Notice we are told that maleic acid has two pKas, thus this is a diprotic acid. For a diprotic acid we can titrate the first proton or both. So we need to determine how many moles of acid and base we have from the information given.

mol Maleic acid = mass / MW = 0.960 g/116.072 gmol⁻¹

= 0.0083 mol

mol KOH = V (Lts) Conc M = (56 mL x 1 L/1000 mL) x 0.295 molL⁻¹

= 0.0165

Now since we have twice as many moles of KOH as maleic acid, we know all the acid will be consumed and the salt of the weak maleic acid, potassium maleate, will be produced.

To determine its pH we will find the concentration of this salt, and since we have the pKa we can determine the pOH and hence pH.

This is so because the maleate anion is the conjugate base of the weak acid maleic acid.

[ maleate anion ] = 0.0083 mol / (0.100 L + 0.056 L) = 0.053 M

Now as usual for weak acids or base, lets setup our ICE table for the hydrolysis reaction:

maleate anion + H₂O ⇆ hydrogen maleate anion + OH⁻

maleate anion hydrogen maleate anion + OH⁻

Initial 0.053 0 0

Change - x + x + x

equilibrium 0.053 - x x x

kw = pka x pkb

pkb = 10¹⁴ / pka

Ka = 10^-pka = 10^-6.27 = 5.37 x 10⁻⁷

Kb = 10⁻¹⁴ / 5.37 x 10⁻⁷ = 1.86 x 10⁻⁸

But Kb = [hydrogen maleate anion] x [OH⁻] / [maleate anion]

Making the approximation 0.053 - x = 0.053 which is reasonable since Kb is very small:

1.86 x 10⁻⁸ = x² / 0.053 ⇒ x = √ (1.86 x 10⁻⁸ x 0.053) = 3.14 x 10⁻⁵

Therefore since x = [OH⁻], the pOH = - log (3.14 x 10⁻⁵) = 4.5

pH = 14 - 4.5 = 9.5