A) A 100.0 mL solution containing 0.8162 g of maleic acid (MW = 116.072 g/mol) i
ID: 713563 • Letter: A
Question
A) A 100.0 mL solution containing 0.8162 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3125 M KOH. Calculate the pH of the solution after the addition of 45.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH?
B) 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.
M2- = ?
HM- = ?
H2M = ?
Explanation / Answer
A) H2M, HM- and M2- are the three forms of maleic acid and the acid ionizations are given as
H2M (aq) --------> H+ (aq) + HM- (aq); pKa1 = 1.92
HM- (aq) --------> H+ (aq) + M2- (aq); pKa2 = 6.27
Millimoles of maleic acid, H2M added = (0.8162 g)/(116.072 g/mol)*(1000 mmole/1 mole) = 7.0318 mmole.
NaOH reacts with H2M as below.
H2M (aq) + NaOH (aq) --------> NaHM (aq) + H2O (l)
As per the stoichiometric equation,
1 mole H2M = 1 mole NaOH = 1 moles NaHM.
Millimoles of NaOH added = (45.00 mL)*(0.3125 M) = 14.0625 mmole.
H2M is completely neutralized to produce 7.0318 mole NaHM. 7.0318 mmole H2M will produce 7.0318 mmole NaHM which reacts further with (14.0625 – 7.0318) mmole = 7.0307 mmole NaOH as below.
NaOH (aq) + NaHM (aq) -------> Na2M (aq) + H2O (l)
7.0307 mmole NaOH reacts with 7.0318 mmole NaHM to give 7.0307 mmole Na2M. The amount of unreacted NaHM is (7.0318 – 7.0307) mmole = 0.0011 mmole.
The total volume of the solution is (100.0 + 45.00) mL = 145 mL.
Determine the equilibrium concentrations as
[HM-] = (0.0011 mmole)/(145 mL) = (0.0011/145) M = 7.5862*10-6 M.
[M2-] = (7.0307 mmole)/(145 mL) = (7.0307/145) M = 0.0485 M.
Use the Henderson-Hasslebach equation as
pH = pKa2 + log [M2-]/[HM-] = 6.27 + log (0.0485 M)/(7.5862*10-6 M)
= 6.27 + 3.8057 = 10.0757 10.07 (ans).
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