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A.) A 20.0 mL sample of 0.0600 M carbonic acid (H2​CO3​) is titrated with 5.00 mL of 0.0960 M KOH...

A.) A 20.0 mL sample of 0.0600 M carbonic acid (H2​CO3​) is titrated with 5.00 mL of 0.0960 M KOH. What is the solution pH to the nearest hundredths place? pKa1​ = 6.35 and pKa2​ = 10.33.

B.) A 20.0 mL sample of 0.0600 M carbonic acid (H2​CO3​) is titrated with 0.0960 M KOH. How many mL of titrant are required to reach the first equivalence point?

C. )A 20.0 mL sample of 0.0600 M carbonic acid (H2​CO3​) is titrated with 0.0960 M KOH. To the nearest hundredth, what is the pH at the first equivalence point?

D.) A 20.0 mL sample of 0.0600 M carbonic acid (H2​CO3​) is titrated with 15.0 mL of 0.0960 M KOH. To the nearest hundredths, what is the pH of the solution? pKa1​ = 6.35 and pKa2​ = 10.33

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Answer #1

pKa1​ = 6.35 and   pKa2​ = 10.33.

A.)

millimoles of H2CO3 = 20.0 x 0.060 = 1.2

millimoles of KOH = 5.00 x 0.0960 = 0.48

H2CO3 +   KOH    -----------> HCO3- +   H2O

1.2              0.48                          0             0

0.72             0                            0.48

pH = pKa1 + log [HCO3- / H2CO3]

    = 6.35 + log [0.48 / 0.72]

pH = 6.17

B.)

At first equivalence point :

mmoles of H2CO3 = mmoles of KOH

20.0 x 0.0600 = 0.0960 x V

V = 12.5 mL

volume of titrant = 12.5 mL

C)

At first equivalence point :

pH = pKa1 + pKa2 / 2

     = 6.35 + 10.33 / 2

pH = 8.34

D)

mmoles of KOH = 15.0 x 0.0960 = 1.44

H2CO3 +   KOH    -----------> HCO3- +   H2O

1.2              1.44                          0             0

0                0.24                       1.2

HCO3-   + KOH   -----------> CO32- +   H2O

1.2             0.24                          0              0

0.96             0                           0.24

pH = 10.33 + log [0.24/ 0.96]

pH = 9.73

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