Question

Oxalic acid is a dicarboxylic acid with structure HOOCCOOH. with K21 = 5.37 x 103 and K 2 = 5.42 105. Consider the titration of 40.0 mL of 0.1 M oxalic acid by 0.1M NaOH and answer the following questions. a) What is the maximum number of protons that can oxalic acid ionize (per molecule)? b) Calculate the pH after the following total volumes of NaOH have been added. (Correct to 2 decimal places.) No marks will be given if the number of decimal places is wrong. i. 0.0 mL of NaOH ii. 20.0 mL of NaOH iii. 40.0 mL of NaOH iv. 60.0 mL of NaOH v. 80.0 mL of NaOH vi. 85.0 mL of NaOH (Take pKw = 14.00)

Answer 1

a) Maximum no. of protons that can oxalic acid ionizes is 2

b) i) 3.067

iI) 4.56

iii) 4.56

iv) 4.56

v) 4.56

vi) 4.56

Volume of oxalic acid = 40 mL

Concentration of oxalic acid =0.1 M

Volume of NaOH = 0 mL

pH= pKa2 +1/2 logC

     = -log (5.42 x10-5) + 1/2 log (0.004)

      = 4.266 -1.199

      = 3.067

Volume of NaOH = 20 mL

Concentration of NaOH =0.1 M

Oxalic acid reacts with 2 moles of NaOH

NaOH    +   H2C2O4             C2O4Na2     + H2O

                    Acid                 Conjugate base

I   2mM             4mM                      -

C 2X                  X                     X

E 2-2X             4-X                    X

Ka = (Ka1 +Ka2)/2     = (5.37 x10-3 + 5.42 x10-5)/2

                                    = 0.00271

Ka =       X/(2-2X)(4-X)

0.00271 = X/(8 + 2X2 -10X)

0.0054X2-0.027X +0.0216 = X

0.0054 X2 -1.027 X + 0.0216 =0

X2 -19.02 X + 4 = 0 (19.02 -18.59)/2

X = 0.215

pH= pKa2 +1/2 logC

     = -log(5.42 x10-5) + 1/2 log (3.785)

     = 4.266 + 0.2890

     = 4.56

Volume of NaOH = 40 mL

Concentration of NaOH =0.1 M

Oxalic acid reacts with 2 moles of NaOH

NaOH    +   H2C2O4             C2O4Na2     + H2O

                    Acid                 Conjugate base

I   4mM             4mM                      -

C 2X                  X                     X

E 4-2X             4-X                    X

Ka = (Ka1 +Ka2)/2     = (5.37 x10-3 + 5.42 x10-5)/2

                                    = 0.00271

Ka =       X/(4-2X)(4-X)

0.00271 = X/(16 + 2X2 -12X)

0.0054X2-0.032X +0.0434 = X

0.0054 X2 -1.032 X + 0.0434 =0

X2 -191.11 X + 8.037 = 0 (191.11 -191.03)/2

X = 0.04

pH= pKa2 +1/2 logC

     = -log(5.42 x10-5) + 1/2 log (3.96)

     = 4.266 + 0.299

     = 4.56

Volume of NaOH = 60 mL

Concentration of NaOH =0.1 M

Oxalic acid reacts with 2 moles of NaOH

NaOH    +   H2C2O4             C2O4Na2     + H2O

                    Acid                 Conjugate base

I   6mM             4mM                      -

C 2X                  X                     X

E 6-2X             4-X                    X

Ka = (Ka1 +Ka2)/2     = (5.37 x10-3 + 5.42 x10-5)/2

                                    = 0.00271

Ka =       X/(6-2X)(4-X)

0.00271 = X/(24 + 2X2 -14X)

0.0054X2-0.038X +0.065 = X

0.0054 X2 -1.038 X + 0.065 =0

X2 -192.22 X + 12.03 = 0 (192.22 -192.09)/2

X = 0.065

pH= pKa2 +1/2 logC

     = -log(5.42 x10-5) + 1/2 log (3.935)

     = 4.266 + 0.297

     = 4.56

Volume of NaOH = 80 mL

Concentration of NaOH =0.1 M

pH = 4.56

Volume of NaOH = 80 mL

Concentration of NaOH =0.1 M

pH = 4.56

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