Question

How many milliliters of 1.27 M KOH should be added to 100. mL of solution containing 10.0 g of histidine hydrochloride (His·HCl, FM 191.62) to get a pH of 9.30?

Answer 1

In histidine hydrochloride is a monoprotic acid.

The reaction between them is : H₃His²⁺ H₂His ⁺ HHis

In this reaction histidine hydrochloride is the intermediate form (H₂His⁺) between the two steps.

For this (1) Must add enough KOH (1:1 molar ratio) to convert all H₂His⁺ to HHis

             (2) Must added more KOH to obtain mixture of HHis and His^- to obtain pH of 9.30

Initial moles of H₂His⁺ is = mass / molar mass

                                       = 10.0 g / 191.62 (g/mol)

                                       = 0.0522 mol

                                        HHis + OH^-        His^- + H₂O

initial moles                   0.0522       a               0

change                              -a          -a              +a

final moles                0.0522-a        -                 a

pH = pKa + log [[His^- ] / [HHis] ]

9.30 = 9.28 + log [a / ( 0.0522 - a ) ]

log[a / ( 0.0522 - a ) ] = 0.02

    a / ( 0.0522 - a ) = 10 0.02

                                = 1.047

    a = 0.055 - 1.047a

2.047 a = 0.055

            a = 0.027 mol

So total number of moles of 0.027 mol + 0.0522 mol = 0.079 mol

So volume of KOH required in L , V = number of moles / Molarity of KOH

                                                         = 0.079 mol / 1.27 M

                                                         = 0.062 L

                                                         = 0.0624 L x 1000 mL /L

                                                         = 62.4 mL

Therefore the volume of KOH required is 62.4 mL