Question

What volume, in liters, of 1.04 M KOH solution should be added to a 0.107 L solution containing 9.26 g of glutamic acid hydrochloride (Glu, FW = 183.59 g/mol; pKa1 = 2.23, pKa2 = 4.42, pKa3 = 9.95) to get to pH 10.37?

Answer 1

moles of glutamic acid hydrochloride, C5H10ClNO4 = mass / molar mass = 9.26 g / 183.59 g/mol = 0.05048 mol

Here the pH required is quite higher than the pKa3 value.

A pH equals to 9.95 will be achieved at half of the third equivalence point.

Hence moles of KOH required to achieve 2nd equivalence point = 2 x 0.05048 mol = 0.101 mol KOH

Now applying Hendersen equation

pH = 10.37 = pKa + log [C₅H₇ClNO₄^3-] / [C₅H₈ClNO₄^2-]

=> 10.37 = 9.95 + log [C₅H₇ClNO₄^3-] / [C₅H₈ClNO₄^2-]

=>[C₅H₇ClNO₄^3-] / [C₅H₈ClNO₄^2-] = 10^0.42 = 2.63

=> moles of C₅H₇ClNO₄^3- / moles of C₅H₈ClNO₄^2-  = 2.63

=> y mol C₅H₇ClNO₄^3- / (0.05 - y) mol C₅H₈ClNO₄^2-  = 2.63

=> y = 0.1315 - 2.63y

=> 3.63 y = 0.1315

=> y = 0.0362 mol

Hence total moles of base required = 0.101 mol KOH + 0.0362 mol KOH = 0.137 mol KOH

Molarity of KOH, M = 1.04 M

=> M x V(L) = 0.137 mol

=> 1.04 mol/L x V(L) = 0.137 mol

=> V(L) = 0.137 mol / 1.04 mol/L = 0.132 L (answer)