HOW DO YOU SOLVE PART 3 I KEPT GETTING IT WRONG The next three (3) problems deal
ID: 1084002 • Letter: H
Question
HOW DO YOU SOLVE PART 3 I KEPT GETTING IT WRONG
The next three (3) problems deal with the titration of 361 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 1.9 M NaOH.
What is the pH of the solution at the 2nd equivalence point? 11.88
What will the pH of the solution be when 0.1266 L of 1.9 M NaOH are added to the 361 mL of 0.501 M carbonic acid? 9.944
How many mL of the 1.9 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 5.935?
Your receipt no. is 168-237 Previous Tries
Explanation / Answer
H2CO3 millimoles = 361 x 0.501 = 181
millimoles of NaOH = 1.9 V
pKa1 = -log 4.3 x 10-7
pKa1 = 6.37
H2CO3 + NaOH ------------------------> NaHCO3 + H2O
181 1.9 V 0 0 ---------------> I
181-1.9V 0 1.9V 1.9V -------------->E
pH = pKa1 + log [NaHCO3 / H2CO3]
5.935 = 6.366 + log (1.9 V / 181-1.9V)
(1.9 V / 181-1.9V) = 0.371
1.9 V = 67.09 - 0.7049 V
2.605 V = 67.09
V = 25.8 mL
volume of NaOH = 25.8 mL
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