The pH of \(0.1 M\) solution of \(NaHCO_3\) (Given \(pK_1 = 6.38 \) and \(pK_2 =10.32 \) ) is

- \(8.35\)
- \(6.5\)
- \(4.3\)
- \(3.94\)

I have doubt in this question. Please type your method in solution.

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## Comments

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TopNewestThe only problem in solving this question is that \(NaHCO_{3}\) (basically, the carbonate ion \(HCO_3^-\)) is amphiprotic (can donate or accept \(H^+\) ion).

So, the total concentration of protons in the water due to the addition of \(NaHCO_3\) will be equal to the number produced, minus the number lost.

Using this and then approximating yields,

\(pH\) = \(\frac{1}{2}\)(\( pK_1 + pK_2\)) = \(\frac{1}{2}\)(\(6.38 + 10.32\)) = \(8.35\).

This implies that the \(pH\) is independent of the concentration.

I saw this solution here.(page 5).

By the way, what score did you get in KVPY?

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Thanks. It is indeed curious that the pH is independent on the concentration itself! Hats off to approximations;)

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Hey,@Siddharth Brahmbhatt , what score did you get in KVPY?

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@Megh Parikh that if we combined my aptitude test score with his interview score, we would have qualified! :D Jokes apart I'm a little disappointed none of us could qualify... 71 is a great score! Had you finished the course earlier? Because the course contained many topics we were yet to learn. I'd really like to know as I might try again this year.

Well I got 59.5 in the aptitude test but only 45.33 in the interview (yeah,it went really bad) thus ending up at 55.96. I just toldLog in to reply

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I didn't understand properly.

What is \(pK_1\) and \(pK_2\)?

I will try to read the PDF.

Aptitude Test Marks out of 100 : 54.5

Interview Marks out of 100 : 65.17

Total Marks : 57.17

What about you?

Congrats on getting qualified for scholarship

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Well, I got 71 in the Aptitude test but in interview I got only 47.5. I don't remember what I did wrong that I got such marks but whatever at least I am selected. How are your studies going on?

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