What volume, in liters, of 1.10 M KOH solution should be added to a 0.117 L solu

Question

What volume, in liters, of 1.10 M KOH solution should be added to a 0.117 L solution containing 10.45 g of glutamic acid hydrochloride (Glu, FW = 183.59 g/mol; pK_a1 = 2.23, pK_a2 = 4.42, pK_a3 = 9.95) to get to pH 10.19? Glutamic acid can act as a buffer at each of its three pK_a values. To solve this problem, you need to take into consideration the amount of KOH that must be added to the glutamic acid solution to move the pH through its first two pK_a values. Therefore, to get Glu into the HGlu^- form (i.e., through H_3Glu^+ and H_2Glu), two (2) moles of KOH must first be added. Then, the problem can be treated as a weak acid/strong base problem, which can be solved using pK_a3 and the Henderson-Hasselbalch equation to determine the moles of OH^- that are required to bring the pH of the glutamic acid solution to 10.19 as a mixture of HGlu^- and Glu^2-.

Explanation / Answer

initial moles of Glu = mass / formula mass = 10.45 / 183.59 = 0.0569

2 moles of OH- should be added per mole of Glu to get the HGlu- form.

moles of KOH = 2 x 0.0569 = 0.1138 mol

HGlu- +   OH- ----------------------> HGlu-2

0.0569 x                                 --------             initial

0.0569-x      0                                     x   -----------------> final moles

pH = pKa + log [Glu-2/HGlu-]

10.19 = 9.95 + log (x / 0.0599 -x)

(x / 0.0599 -x) = 1.738

x = 0.0380

total moles of OH - = 0.038 + 0.1138 = 0.1518

KOH volume = moles of KOH / molarity

                     = 0.1518 / 1.10

                     = 0.138 L

volume of KOH = 0.138 L

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