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The 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.

@Shaan Vaidya
–
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 told @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.

@Siddharth Brahmbhatt
–
Yes, I had finished the course. Of course, all the extra we learnt were just mere basic concepts. We had extra classes scheduled at our tuitions for completing the course. We did that only in Chemistry and Math. In physics, we only practised what we had learnt because KVPY papers even include questions about complex electric circuits which we could never have learnt in such a short time. And yes, I also attended Biology classes specially scheduled for KVPY preparations.(Out of all sections, I got the most marks in Biology! I don't know how! :P). Surely, this time you will crack KVPY. BEST OF LUCK.

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?

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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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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