Quick, a G.Chem problem.

[UCdannyLA]

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If the PKa value for H2CO3 is 6.4, what is the pH of a 10^-3 M solution of this acid?

 

[Drakensoul]

Since it's diprotic, you'd ideally need to do this twice, but since they didn't give the Ka for HCO3-, and because the resultant amount is so small, I'm assuming you're expected to ignore it.

pH = -log[H]

H2CO3 --> HCO3- + H+ @ Ka=6.4

[HCO3-] [H+]
-------------- = 6.4
[H2CO3]

[x][x]
------------ = 6.4
[.001-x]

.0064-6.4x=x^2
x^2 + 6.4x - .0064 = 0

x~.001

-log[.001] = 3

So, 3 would be my guess! I'm way out of practice, though. 🙂

 

[geckoUBC]

pH = 3 ish?

 

[MDtoBe777]

I may be completely off base (I suck at acid/base chem), but shouldn't you use the Ka value, not the pKa value when determining the concentration of [h+] in the solution? So you have the pKa being 6.4, so don't you need to work backward to get the Ka?

And also, you can determine the pH at the equivalence point, can't you? Isn't that where the ph=pka? But, you do need to know that you are at the half equivalence point. So, they would have to either tell you explicitly or provide you with a graph.

PLEASE correct me if any of this is incorrect (which it very well may be). 😕

 

[YML]

Pka is just -log ka

P of anything for that matter is - log of anything

Ph = -log H+ conc
Poh= -log oh conc

 

[bgreet]

Treat it like its a monoprotic acid that completely dissociates. Assume that the second hydrogen will be really hard to have come off, therefore the ph should be approx 3

 

[MDtoBe777]

Here is what I'm thinking:

[H+][HCO3-]/[H2CO3]=Ka

Ka should be around 4.2X10^-7 since pKa=6.4

Then x^2/10^-3=4.2X10^-7
x^2=4.2X10^-10
so x is roughly 10^-5.
-log(x)=ph
-log(10^-5)=5

so my guess is pH 5? 😕

 

[SOUNDMAN]

I think MDtobe777 is correct. Drakensoul you forgot to change the pKa of 6.4 to Ka and that is where your calculation is wrong...however the rest of your concept and how you solved it is correct.

 

[MDtoBe777]

👍

 

[UCdannyLA]

Well, I thought I had lost the explanations, but I found it.

the answer is 4.7.

PKa = 6.4, which means Ka = 4 x 10^-7

Since Ka is hte dissociation constant,

Ka = [x]* [x] / (.001 M)

4 x 10^-7 = x^2 / (.001 M)

[x] = 2 x 10^-5

pH = -log [x]
pH = -log [2 x 10^-5]
pH = 4.7


Please tell me that this kind of tedious question won't appear on the MCAT...! is there a short cut?

 

[willthatsall]

Well, I thought I had lost the explanations, but I found it.

the answer is 4.7.

PKa = 6.4, which means Ka = 4 x 10^-7

Since Ka is hte dissociation constant,

Ka = [x]* [x] / (.001 M)

4 x 10^-7 = x^2 / (.001 M)

[x] = 2 x 10^-5

pH = -log [x]
pH = -log [2 x 10^-5]
pH = 4.7


Please tell me that this kind of tedious question won't appear on the MCAT...! is there a short cut?

Nope, not really a shortcut that I know of, but I have seen problems like this on Kaplan stuff. The only shortcut is to be good at converting logs.

 

[ntc1983]

For weak reagents, you can use this equation: pH = 0.5 pKa - 0.5log[HA] for weak acids, and pOH = 0.5pKb - 0.5log[A-] for weak bases.

 

[Kownell]

MDtobe has the procedure correct.

I got the same pH at around 4.5--5.

Keep the questions coming.

 

[Drakensoul]

Oh, I didn't even notice that he said pKa 😀
I called it Ka when I solved it.. woops 🙂

4.7 it is!

 

[geckoUBC]

This might be a dumb question, but... if pKa is 6.4, would Ka not be 10^-6.4?
Where are you guys getting these other values for Ka?

 

[Drakensoul]

This might be a dumb question, but... if pKa is 6.4, would Ka not be 10^-6.4?
Where are you guys getting these other values for Ka?

10^-6.4 ~ 4x10-7

 

[Doctor Bagel]

and all i can say is that it is so much easier for me to solve that problem keeping is 10 ^ -6.4.

 

[shaq786]

CAN SOMEONE WALK ME THROUGH STEP BY STEP ON HOW TO DO THE CALCULATIONS BY HAND????

FOR EXAMPLE...how do you take the square root of x^2 = 10^-9.4????

 

[PRamos]

Please tell me that this kind of tedious question won't appear on the MCAT...! is there a short cut?

There is an easy way to get pH of a weak acid in the BR general chemistry book. That is what ntc1983 used.

For weak reagents, you can use this equation:

• pH = 0.5 pKa - 0.5log[HA] for weak acids, and
  pOH = 0.5pKb - 0.5log[A-] for weak bases.

It is an awesome short cut that makes questions like this easy.

pH = 1/2(6.4) - 1/2log[0.001] = 3.2 - 1/2(-3) = 3.2 + 3/2 = 3.2 + 1.5 = 4.7

That is SOOO much easier than the method you wrote.

This is a perfect example of why I love the BR materials more than the PR materials I used the first time. I remember struggling with that question the first time I studied. Acid and Base stuff is so easy the they taught us to do it.

 

[DieselPetrolGrl]

but um i didnt get the right answer when i kept it 10^-6.4....like i cant dare to aprox logs at all so i wouldnt have deduced 4x10^-7

i got 3
1x10^-7 = x^2/1x10^-3
x^2= 1x10^-10
x=1x10^apx-3
ph=-log 1x10^apx-3
ph=3

🙁🙁🙁

 

[DieselPetrolGrl]

There is an easy way to get pH of a weak acid in the BR general chemistry book. That is what ntc1983 used.



It is an awesome short cut that makes questions like this easy.

pH = 1/2(6.4) - 1/2log[0.001] = 3.2 - 1/2(-3) = 3.2 + 3/2 = 3.2 + 1.5 = 4.7

That is SOOO much easier than the method you wrote.

This is a perfect example of why I love the BR materials more than the PR materials I used the first time. I remember struggling with that question the first time I studied. Acid and Base stuff is so easy the they taught us to do it.

that only works for weak acids and bases...do u mean anything that protonates only1x (ie : only 1 H dissassociates in this case)

thanks!

 

[Doctor Bagel]

dieselpetrolgirl,
i think you did the exponent wrong. to get the square root of an exponent, you divide by two, so the square root of 1 x 10^-10 is 1 x 10^-5. basically, you're raising each side to the 1/2. i did that, too, initially, and got a similar answer to yours.

 

[DieselPetrolGrl]

dieselpetrolgirl,
i think you did the exponent wrong. to get the square root of an exponent, you divide by two, so the square root of 1 x 10^-10 is 1 x 10^-5. basically, you're raising each side to the 1/2. i did that, too, initially, and got a similar answer to yours.

omg thank you!
like i admitted - im a goof with exponents i sooo miss my calculator!
but i see my error -5 exp x -5 exp = -10 * sheepish grin * 😛

🙂 thank you again! ! ! 🙂