Part A Consider the titration of 25.0 mL of 0.0200 M H2 CO3 with 0.0250 M KOH. C

Question

Part A Consider the titration of 25.0 mL of 0.0200 M H2 CO3 with 0.0250 M KOH. Calculate the pH after the addition of 10.0 mL of base. (Ka, 4.3 × 10-7 and Ka,-5.6 × 10-11 .) Express your answer using two decimal places. ? pH4.56 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part B Consider the titration of 25.0 mL of 0.0200 M H2CO3 with 0.0250 M KOH. Calculate the pH after the addition of 20.0 mL of base. (Ka, 4.3 × 10-7 and Ka,-5.6 × 10-11 .) Express your answer using two decimal places. pH =

Explanation / Answer

Part A:

Moles H2CO3 = 0.0250 L x 0.0200 M = 0.0005
moles KOH = 0.0100 L x 0.025 M = 0.00025
H2CO3 + KOH HCO3- + H2O
moles H2CO3 = 0.000500 - 0.00025 = 0.00025
moles HCO3- = 0.00025
pKa1 = - log[4.3 x10-7] = 6.36
pH = 6.36 + log 0.00025/0.00025 = 6.36

Part B:

Moles H2CO3 = 0.0250 L x 0.0200 M = 0.0005
moles KOH = 0.0200 L x 0.025 M = 0.0005

Total volume = 0.045 L

[HCO3-] = 0.0005 M/0.045 L = 0.0111 M
HCO3- H++ CO32-

Now the complication comes in because HCO3- can not only pick up protons to reform H2CO3 but it can also give off protons to form CO32-.
We get the net reaction
2HCO3- H2CO3+ CO32-
The K for HCO3- H++ CO32- is just Ka2
The K for HCO3- + H+ H2CO3 is 1 /Ka1

consequently for the
2HCO3- H2CO3+ CO32-
K = Ka2 x (1 /Ka1) = (5.6x10-11­) x (1/4.3 x10-7) = 1.302x10-4
Let 2x mol/L HCO3- react by the above net reaction.
This will form x mol/L CO32- and x mol/L H2CO3, while reducing the HCO3- concentration from 0.0111 to 0.0111 -2x

1.302x10-4 = x2 / (0.0111 -2x)2
the equation can be solved by taking the square root of each side.
0.0141 = x / 0.0111-2x

0.000156 - 0.0282 x = x

0.000156 = 1. 0282 x
x = 1.517x10-4 M

[H2CO3] = [CO32-]= 1.517x10-4 M

[HCO3-] = 0.0111 - 2 ( 0.0001517)= 0.01079 M

now we can calculate [H+] by substituting these values in Ka1
4.3 x10-7= [H+] x 0.01079 / 0.0001517

[H+] = 0.0604 x10-7M
pH = -log[0.0604 x10-7] = 8.21

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