Question

40.00 mL of a 0.1000 M carbonic acid (H₂CO₃) was titrated with 0.2000 M sodium hydroxide.

carbonic acid pka1:6.35 pka2:10.33

1.Calculate the volume of sodium hydroxide required to reach the first equivalence point.

2.Calculate the volume of sodium hydroxide required to reach the second equivalence point.

3.Write the balanced acid-base reaction and calculate the pH before the addition of sodium hydroxide.

4.Write the balanced acid-base reaction and calculate the pH after the addition of 12.00 mL of sodium hydroxide

5.Write the balanced acid-base reaction and calculate the pH after the addition of 20.00 mL of sodium hydroxide

6. Write the balanced acid-base reaction and calculate the pH after the addition of 30.00 mL of sodium hydroxide

7.Write the balanced acid-base reaction and calculate the pH after the addition of 40.00 mL of sodium hydroxide

8.Write the balanced acid-base reaction (paper only) and calculate the pH after the addition of 44.00 mL of sodium hydroxide

Answer 1

1.Calculate the volume of sodium hydroxide required to reach the first equivalence point.

volume of NaOH   = 40 x 0.100 / 0.2 = 20 mL

2.Calculate the volume of sodium hydroxide required to reach the second equivalence point.

volume of NaOH = 40 x 0.100 x 2 / 0.2 = 40 mL

3.Write the balanced acid-base reaction and calculate the pH before the addition of sodium hydroxide.

H2CO3 --------------------> H+ + HCO3-

pH = 1/2 [pKa1 -log C]

pH = 1/2 [6.35 -log 0.1]

pH = 3.68

4.Write the balanced acid-base reaction and calculate the pH after the addition of 12.00 mL of sodium hydroxide

mmoles of H2CO3 = 40 x 0.1 = 4

mmoles of NaOH = 12 x 0.2 = 2.4

H2CO3   +   NaOH -------------------> NaHCO3 + H2O

4                  2.4                                 0               0

1.6                0                                  2.4            --

pH = pKa + log [NaHCO3 / H2CO3]

pH = 6.35 + log (2.4 / 1.6 )

pH = 6.53

5.Write the balanced acid-base reaction and calculate the pH after the addition of 20.00 mL of sodium hydroxide

mmoles of H2CO3 = 40 x 0.1 = 4

mmoles of NaOH = 20 x 0.2 = 4

H2CO3   +   NaOH -------------------> NaHCO3 + H2O

4                  4                               0               0

0                0                                 4            --

here only NaHCO3 is left . so pH depends on pKa and pKa2

pH = 1/2 [pKa1 + pKa2 ]

pH = 1/2 [ 6.35 + 10.33]

pH = 8.34

6. Write the balanced acid-base reaction and calculate the pH after the addition of 30.00 mL of sodium hydroxide

mmoles of H2CO3 = 40 x 0.1 = 4

mmoles of NaOH = 30 x 0.2 = 6

H2CO3   +   NaOH -------------------> NaHCO3 + H2O

4                  6                                0               0

0                2                              4            --

NaHCO3   + NaOH -----------------------> Na2CO3 + H2O

4                       2                                           0               0

2                      0                                         2                  2

pH = pKa2 + log [Na2CO3 / NaHCO3]

pH = 10.33 + log (2 /2 )

pH = 10.33

7.Write the balanced acid-base reaction and calculate the pH after the addition of 40.00 mL of sodium hydroxide

mmoles of NaOH = 40 x 0.2 = 8

balanced equation : H2CO3 + 2 NaOH ------------------> Na2CO3 + H2O

here only Na2CO3 is left

Na2CO3 concentration = 4 / (40 + 40) = 0.05 M

pKb1 = 3.67

pOH = 1/2 [pKb -log C]

pOH = 1/2 [3.67 -log 0.05]

pOH = 2.48

pH = 11.5

8.Write the balanced acid-base reaction (paper only) and calculate the pH after the addition of 44.00 mL of sodium hydroxide

balanced equation : H2CO3 + 2 NaOH ------------------> Na2CO3 + H2O

pH = 11.98