Question

1. Tartaric acid is added to foods as an antioxidant and to impart its distinctive sour taste. A chemist prepares a solution that contains a formal concentration of 0.05 M. Phthalic Acid: pKa1 (HOOCCH(OH)CH(OH)COOH/HOOCCH(OH)CH(OH)COO-) = 2.89, pKa2 (HOOCCH(OH)CH(OH)COO-/-OOCCH(OH)CH(OH)COO-)) = 4.40,

a) Calculate the pH of the solution.

b) The chemist then takes 100.00 mL of the tartaric acid solution and adds enough NaOH (0.10 M) solution to change the pH to pH=4.00. Calculate the fraction of each of the species in the solution at pH 4.00.

c) Plot an approximate curve of versus pH for each species.

Answer 1

a) pka2<7 or neutral pH of water. So,it will get deprotonted on hydrolysis

Pthalic acid dissociates according to the equation,

HOOCCH(OH)CH(OH)COOH + 2H2O -OOCCH(OH)CH(OH)COO^- + 2H3O⁺

Concentration of pthalic acid=0.05M

pka2=4.40=-log ka2

ka2=10^-pka2 =3.981*10^-5

ka2=[-OOCCH(OH)CH(OH)COO-][H3O+]^2/[HOOCCH(OH)CH(OH)COOH]

ICE tble:

                       [HOOCCH(OH)CH(OH)COOH]               [-OOCCH(OH)CH(OH)COO-]                 [H3O+]

initial                      0.05M                                                      0                                              0

change                      -x                                                         +x                                             +2x

equilibrium                0.05M-x                                                  x                                              2 x

ka2=(x)(2x)/0.05-x=2x^2/0.05 [0.05>>>x as very less dissociation takes place,due to very small ka2]

3.981*10^-5=2x^2/0.05

or, 9.953*10^-7=x

So,[H3O+]=2x=2*9.953*10^-7=1.990*10^-6M

pH=-log [H3O+]=-log 1.990*10^-6M=5.7

pH=5.7

b)mol of tartaric acid taken=0.05 mol/L*100ml=0.05 mol/L*0.1L=0.005 mol

Using Henderson-hasselbach equation,

pH=pka+log [base]/[acid]

or, 4.00=pka2+log [-OOCCH(OH)CH(OH)COO-]/[HOOCCH(OH)CH(OH)COOH]

or, 4.00=4.4 +log [-OOCCH(OH)CH(OH)COO-]/[HOOCCH(OH)CH(OH)COOH]

So, [-OOCCH(OH)CH(OH)COO-]/[HOOCCH(OH)CH(OH)COOH]=2.512:1

[-OOCCH(OH)CH(OH)COO-]=(2.512/2.215+1)*0.05M=0.0358M

[HOOCCH(OH)CH(OH)COOH]=(1/2.215+1)*0.05M=0.0155M

fraction, of two species in solution -OOCCH(OH)CH(OH)COO- and HOOCCH(OH)CH(OH)COOH is 2.512:1