Two equations that I have difficulty doing in my head. Tips please?

[ProteinChemist]

#1
For calculating proton conc from pKa the equation starts as Ka=10^-pKa

So say you get Ka=10^-3.85

How do I do that in my head to get it to 1.41x10^-4??

#2
For calculating the K for a free energy problem, the equation is:

K=e^-dG/RT

So I get K=e^37 but how do I find a number for this w/o using a calculator?

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[BerkReviewTeach]

#1
For calculating proton conc from pKa the equation starts as Ka=10^-pKa

So say you get Ka=10^-3.85

How do I do that in my head to get it to 1.41x10^-4?

I'd suggest working backwards and then approximating. We teach that:

-log (1.41 x 10^-4) = -(-4) - log 1.41 = 4 - log 1.41

log 2 = 0.3 and root 2 = 1.4, so log 1.41 = 0.15

This means that -log (1.41 x 10^-4) = 4 - 0.15 = 3.85

• Use the shortcut -log (n x 10^-#) = -(-#) - log n

For instance, if Ka = 3 x 10^-5, then pKa = 5 - log 3 = 4.5 or so.

If [H+] = 5 x 10^-3, then pH = 3 - log 5 = 2.3 or so.

If [OH-] = 2 x 10^-6, then pOH = 6 - log 2 = 5.7 and pH = 9.3.

#2
For calculating the K for a free energy problem, the equation is:

K=e^-dG/RT

So I get K=e^37 but how do I find a number for this w/o using a calculator?

First off, what are you studying that you actually think you'd have to do that? That said, here is a reasonable trick.

2.301 log X = ln X

as such, e^x = 10^(x/2.3)

e^37 = 10^(37/2.3) which roughly equals 10^16 (which hopefully a close enough answer for a multiple choice question).

 

[FutureScaresMe]

log 2 = 0.3 and root 2 = 1.4, so log 1.41 = 0.15

How did you figure that part out?

Edit: Ohhh

log(2) = 0.3 so 10^0.3 = 2
2^0.5 = 1.4
log(1.4) = x so 10^x = 1.4

10^x = (10^0.3)^0.5

squaring both sides:

2x = 0.3
x= 0.15 ... clever

 

[ProteinChemist]

Thanks much!

I'm using REA (Research and Education Association "The Best Test Preparation for the MCAT".

These questions are from a practice test:

What is the equilibrium constant for the reaction of glucose going to pyruvate at STP if the dG for the reaction is -17.5kcal/mol?

Pyruvic acid dissociates in water to form pyruvate anion. The pKa of pyruvic acid is 3.85. What concentration of pyruvate ion is present when 0.5 mole of pyruvic acid dissociates in 1L of water?

The pyruvate conc is very sensitive to pH. What is the pyruvate conc if 0.5 moles of pyruvate are dissolved in 1L of water that has an H+ conc of 0.1M?

Is there an easier way to do these questions than using the equations I had? Or are these questions really still on the MCAT?

Honestly, it's a lot of calculation and I figured I'd just skip these the first time I went over the test and come back to it if I had time.

 

[BerkReviewTeach]

Thanks much!

I'm using REA (Research and Education Association "The Best Test Preparation for the MCAT".

These questions are from a practice test:

What is the equilibrium constant for the reaction of glucose going to pyruvate at STP if the dG for the reaction is -17.5kcal/mol?

Pyruvic acid dissociates in water to form pyruvate anion. The pKa of pyruvic acid is 3.85. What concentration of pyruvate ion is present when 0.5 mole of pyruvic acid dissociates in 1L of water?

The pyruvate conc is very sensitive to pH. What is the pyruvate conc if 0.5 moles of pyruvate are dissolved in 1L of water that has an H+ conc of 0.1M?

Is there an easier way to do these questions than using the equations I had? Or are these questions really still on the MCAT?

Honestly, it's a lot of calculation and I figured I'd just skip these the first time I went over the test and come back to it if I had time.

Just a subtle suggestiojn, but you might want to use absolutely anything else on the market except REA. The questions they use are modifcations of what they used back in the 1980s. Crazy math questions like the second one you asked are no longer a part of the MCAT experience. They definitely have math questions, but they are easier to do using approximatiosn than that example.

 

[ProteinChemist]

Oy vey! My friend gave it to me thinking she was doing me a favor.

Thanks so much for saving me the trouble. Questions from the 1980s? Shhhhhit.

I'll get myself a newer book. Thank you thank you thank you.

 

[BerkReviewTeach]

Oy vey! My friend gave it to me thinking she was doing me a favor.

Thanks so much for saving me the trouble. Questions from the 1980s? Shhhhhit.

I'll get myself a newer book. Thank you thank you thank you.

Start with the classifieds here at SDN for an idea of what is available and roughly how much it goes for used. There are some bargains.

 

[futuredoctor10]

Use the shortcut -log (n x 10^-#) = -(-#) - log n

This is a good shortcut. If answer choices are not close to each other (which is often the case), another method I use, credit to EK General Chemistry:

pH = - log H+

If H+ = [1.41 x 10^-4], what is the pH?
pH = -log 1.41 x 10^-4

pH will be between the exponent value (4) and one number lower (3).
Because it is 1.41, it will be closer to 4.

if Ka = 3 x 10^-5, then pKa is between 4 and 5.

If [H+] = 5 x 10^-3, then pH is between 2 and 3.

If [OH-] = 2 x 10^-6, then pOH is between 5 and 6. pH is between 9 and 8.