Question

Calculate the volume, in liters, of 1.616 M KOH that must be added to a 0.129 L solution containing 10.85 g of glutamic acid hydrochloride (H3Glu Cl–, MW = 183.59 g/mol) to achieve a pH of 10.17. Glutamic acid (Glu) is an amino acid with pKa values of pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95.

=?? L

Answer 1

First covert your grams of Glu to moles of Glu:

Number of moles / molar mass = 10.85 g/ 183.59 g/ mole

= 0.0590 moles

The reactions between Glu and KOH are as follows:

H3Glu+ + OH- --->(pKa1) H2Glu + H2O
H2Glu + OH- --->(pKa2) HGlu- + H2O
HGlu- + OH- --->(pKa3) Glu2- + H2O

here; the first 2 reactions use 2 moles of KOH. So you need the moles of KOH in the ICE table to match the moles of H3Glu+ and H2Glu.

2*(0.0590 mol KOH) =0.118 mol KOH

to calculate the number of mole of KOH, the Henderson-Hasselbalch equation, as follows:

HGlu- + OH- ------------> Glu2- + H2O
pKa3
I 0.0590 mol x 0 --
C - x -x +x --
E 0.0590 mol Glu - x 0 x

pH = pKa3 + log (Glu2-/HGlu-)
10. 17 =9.95 + log ( x /0.0590 mol - x)

0.22 = log ( x /0.0590 mol - x)

10^0.22 = x /0.0590 mol – x

1.66 = x /0.0590 mol – x

0.09794 – 1.66 x= x

2.66 x= 0.09794

X= 0.0368 mole
x = mol OH- = 0.0368 mol OH-

Now add all the moles of OH-:
0.118 mol + 0.0368 mol = 0.1548 mol OH-

And find the litres:

Volume in L = number of mole s/ molarity

= 0.1548 mol OH- /1.616 M KOH

= 0.0958 L

= 95.8 ml