Question

Calculate the volume, in liters, of 1.687 M KOH that must be added to a 0.122 L solution containing 10.68 g of glutamic acid hydrochloride (Hg Glu+ciº,MW = 183.59 g/mol) to achieve a pH of 10.31. Glutamic acid (Glu) is an amino acid with pKa values of pKal = 2.23, pKa2 = 4.42, and pKa3 = 9.95. NH, C O HC-CH2-CH2-C-OH C= 0 OH Glutamic Acid Hydrochloride volume:

Answer 1

Concentration of KOH= 1687 M (given) Gram wt. of qutamic aud = 10'68 gm C given) Molecular wt of glutamic au'd- 163.599mmola i No. of moles of glutamic and gm wt - 10.68, Cinital moles) M.W 183.59 = 0.0582 " there are two carboxylic and groups are present in glutamic aud, we need two moles of base to convert it into conjugate boje. . Moles of KOH requind = ax motes of glutamic aud - 2X0.0562 = 0.4464 Now for H Glu toho – Glu? Enital 0.0562 x 0 final 0.0582-x 0 x i from Henderson- Hasselblch egten pH= p ka t log (Gly? ] 1031- 995 + log 10 .82-x) log (6 9 1882-x) = 10.31-9.95 = 0.36 100.36 0.0582x -

0.0580x = 2.290 x= 0.133 278 2.29 0x ir x= 0.133278 3.29 x= 0.041 en total moles of KOH = 0.041+0. 1164 = 0.1574 KOH volume – moles of Koh cone of ROH - 0.1574 1.667 E KOH volume= 0.0934)