The next three (3) problems deal with the titration of 301 mL of 0.501 M carboni

Question

The next three (3) problems deal with the titration of 301 mL of 0.501 M carbonic acid (H2CO3) (Ka1 = 4.3 x 10-7, Ka2 = 5.6 x 10-11) with 2.1 M NaOH.
I HAVE DONE 2 OF THEM, JUST NEED HELP WITH THE LAST ONE. PLEASE HELP.

What is the pH of the solution at the 2nd equivalence point? 11.89

What will the pH of the solution be when 0.09694 L of 2.1 M NaOH are added to the 301 mL of 0.501 M carbonic acid?  9.983

I NEED HELP WITH THIS ONE:

How many mL of the 2.1 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 6.172?

Explanation / Answer

H2CO3+2NaOH => Na2CO3 + 2H2O

the pH of the solution be when 0.09694 L of 2.1 M NaOH are added to the 301 mL of 0.501 M carbonic acid = 9.983

then the moles of NaOH = 2.1*0.09694 = 0.203574 moles

for pH = 6.172 , moles of NaOH required = 0.203574*6.172/9.983 = 0.12586

mL of the 2.1 M NaOH are needed to raise the pH of the carbonic acid solution to a pH of 6.172 = 0.12586/2.1 = 0.0599L = 59.9 mL

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