**** pH calculations, are they necessary?

[pfaction]

TBR, 4.21

What is the pH of 0.07562 M HCO2H with a pKa of 3.642?
a) 2.259
b) 2.383
c) 2.759
d) 2.883

**** THIS QUESTION, every time I do it whether I round up or down I get between answers and I always choose wrong. Do we need to know intricacies of how to do this, because I really hate this ****.

I also got this wrong:

What is the pH of 0.20M sodium propionate, if it has Kb 7.2e-10?

pKb is thus 10-log(7) = 10-(0.85) = 9.2
Therefore, pKa = 14-9.42 = I rounded to 5.
So, pH = pKa + log (2)
pH = 5 + log (0.2)
Log 0.2 = log (2x10^-2) = 2-log2 = 2-0.3 = 1.7
pH = 5 + 1.7 = 6.7

NOPE, THE ANSWER IS BETWEEN 7 AND 11.

**** this.

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[cloak25]

I got it wrong the first time due to careless math.

For the first one, I used pH=1/2pka - 1/2log[HA]

it's all approximations from here. 0.07562 is between 0.1 and 0.01.

ph=0.5(3.642)-0.5log(0.01) = about 2.8

ph=0.5(3.642)-0.5log(0.1) = about 2.3

0.07562 is closer to 0.1 than it is to 0.01 so the answer should be a little greater than 2.3. B is correct?

 

[hellocubed]

pH = pKa + log (2)

Wait, where exactly did you get this formula.
Sorry Im at work right now and I can't confirm.

I thought the formula was
pH= (1/2)pKa - (1/2)Log [HA]

What is the pH of 0.07562 M HCO2H with a pKa of 3.642?
a) 2.259
b) 2.383
c) 2.759
d) 2.883

pH= (1/2) 3.642 - (1/2) Log [0.077]
pH= 1.82-(1/2) Log [0.01] + Log [8]
pH= 1.82-(1/2) Log [0.01] + Log [2] + Log[2] + Log[2]
pH= 1.82- (1/2) (-2+0.3+0.3+0.3)
pH= 1.82 - (1/2) (-1.1)
pH= 1.82 - -.55
pH= 2.37


and because we rounded Log[0.07562] to Log [0.08], the answer should be slight larger.
Hence B

 

[Melomare17]

i thought that kind of calculation was too in depth for us to know? i didnt think they went past 1 or 2 decimal points max

 

[cloak25]

The question was purposely designed that way to test your ability to approximate/use shortcuts.

 

[hellocubed]

they do, but just expect us to round the numbers generously.
We just have to compensate for the rounding.

 

[Class1P]

The actual MCAt choices won't have pHs that are so close together. If they are, double check your math.

 

[pm1]

where does this formula comes from?

pH= (1/2)pKa - (1/2)Log [HA]

its not from Henderson Hasselbach, right?

 

[cloak25]

where does this formula comes from?

pH= (1/2)pKa - (1/2)Log [HA]

its not from Henderson Hasselbach, right?

It's from TBR

 

[pm1]

It's from TBR

but I mean how is this formula derived? thanks

 

[MedPR]

where does this formula comes from?

pH= (1/2)pKa - (1/2)Log [HA]

its not from Henderson Hasselbach, right?

It is from HH iirc. Just memorize it anyway, it's awesome. Wasn't useful on today's MCAT though.

 

[gettheleadout]

I did this the old-fashioned way (the way I would do it on a gen chem test if given one right now) using an ICE table and the Ka and got B without any issues with rounding (though I used a calculator.)

I have never seen that equation from TBR but that looks awesome!

 

[pm1]

I did this the old-fashioned way (the way I would do it on a gen chem test if given one right now) using an ICE table and the Ka and got B without any issues with rounding (though I used a calculator.)

I have never seen that equation from TBR but that looks awesome!

I would also solve it using the ICE table but I am not being able to get to B.
Wouldn't we just have to do Ka = [H+][CO2H-]/[HCO2H]

HCO2H -----------------> H+ + CO2H-
0.07562 0 0
-x +x +x
round to 0.07562 x x

from pKa I found Ka to be approx 4 x 10^-4

4x10^-4 = x^2 / 0.07562
doing this I don't get to the right answer.. what am i doing wrong?
Thank you!!

 

[pfaction]

Can someone walk me through it? I flat out don't understand. Every single step, if possible.

 

[ipodtouch]

What is the pH of 0.20M sodium propionate, if it has Kb 7.2e-10?

pKb is thus 10-log(7) = 10-(0.85) = 9.2
Therefore, pKa = 14-9.42 = I rounded to 5.
So, pH = pKa + log (2)
pH = 5 + log (0.2)
Log 0.2 = log (2x10^-2) = 2-log2 = 2-0.3 = 1.7
pH = 5 + 1.7 = 6.7

NOPE, THE ANSWER IS BETWEEN 7 AND 11.

Could anyone actually take a shot at this problem?
I'm trying, but I cannot get an answer between 7 and 11.

pKa calculations appear to be correct, to be around 5.
But when I apply the calculations with pKa=5 and 0.2M, I get a very low pH.

pH=2.5 - (1/2)log[0.2]
pH=2.5 - (1/2) (-1+0.3)
pH=2.85

 

[gettheleadout]

I would also solve it using the ICE table but I am not being able to get to B.
Wouldn't we just have to do Ka = [H+][CO2H-]/[HCO2H]

HCO2H -----------------> H+ + CO2H-
0.07562 0 0
-x +x +x
round to 0.07562 x x

from pKa I found Ka to be approx 4 x 10^-4

4x10^-4 = x^2 / 0.07562
doing this I don't get to the right answer.. what am i doing wrong?
Thank you!!

Your ICE table is set up correctly, but your calculation of Ka is off. We are given that pKa = 3.642, so since 3.624 = -log(Ka), Ka = 10^(-3.642) = 2.28E-4

 

[MedPR]

Could anyone actually take a shot at this problem?
I'm trying, but I cannot get an answer between 7 and 11.

pKa calculations appear to be correct, to be around 5.
But when I apply the calculations with pKa=5 and 0.2M, I get a very low pH.

pH=2.5 - (1/2)log[0.2]
pH=2.5 - (1/2) (-1+0.3)
pH=2.85

The pKa+pKb=14 is only true for conjugates. In other words, if the pKb for sodium propanoate is 9, the pKa for sodium propanoate is not 14-9.

 

[kasho11]

I think I got it right?

So Kb = [OH-][conj acid]/[base] = 7.2 x 10^-10 = x^2/.2-x

You assume x to be much smaller than .2 so becomes x^2/.2

so turn it into 2 x 10^-1 multiplied by 7.2 x 10^-10 = 14.4 x 10^-11

Turn that into 1.44 x 10^-10 = x^2, 1.44 x 10^-10 perfectly squares out to 1.2 x 10^-5.


-log of that OH concentration = ~ 4.9, pOH. Meaning pH = 14 - 4.9 = 9.1

 

[ipodtouch]

oh nvm, haha

 

[MedPR]

I think I got it right?

So Kb = [OH-][conj acid]/[base] = 7.2 x 10^-10 = x^2/.2-x

You assume x to be much smaller than .2 so becomes x^2/.2

so turn it into 2 x 10^-1 multiplied by 7.2 x 10^-10 = 14.4 x 10^-11

Turn that into 1.44 x 10^-10 = x^2, 1.44 x 10^-10 perfectly squares out to 1.2 x 10^-5.


-log of that OH concentration = ~ 4.9, pOH. Meaning pH = 14 - 4.9 = 9.1

Looks good to me.

 

[gettheleadout]

Can someone walk me through it? I flat out don't understand. Every single step, if possible.

So we begin by setting up an ICE table for the reaction (in this case the deprotonation of our weak acid, HCO2H.) We are given the starting concentration of the acid and can assume no initial presence of H+ or conjugate base. We represent the amount of our acid that will deprotonate as x, and subtract it from the initial concentration in the Change row, simultaneously adding x to each column for the products (as for each molecule of HCO2H that deprotonates, we gain a free H+ and a molecule of CO2H-):

___HCO2H <---> H+ + CO2H-
I__0.07562 M____0_____0___
C____-x________+x____+x__
E__0.07562-x____x_____x___

Given that pKa = 3.642 and pKa = -log(Ka):

3.642 = -log(Ka)

Rearranging: Ka = 10^(-3.642) = 2.28E-4

The equilibrium constant Ka for the dissociation of our weak acid should equal the equilibrium concentrations derived from our ICE table:

Ka = ( [H+]eq[CO2H-]eq ) / [HCO2H]eq

For [HCO2H]eq we can assume the change (-x) is negligible because the ratio of initial concentration to Ka is greater than 100: (0.07562 / 2.28E-4) > 100. [HCO2H]eq is then equal to the initial concentration of HCO2H.

Back to the equilibrium expression...

2.28E-4 =( [x][x] ) / [0.07562] = x^2 / 0.07652
...solving for x...
x = 4.15E-7 = [H+]eq

pH = -log[H+] = -log[4.15E-7] = 2.382

 

[MedPR]

Btw, when trying to follow equations and explanations (like the one above), I found it best to write them out and see them in my own handwriting. I found it almost impossible to follow derivations and explanations in computer text.

 

[gettheleadout]

Btw, when trying to follow equations and explanations (like the one above), I found it best to write them out and see them in my own handwriting. I found it almost impossible to follow derivations and explanations in computer text.

I think this is good advice and would like to add that it's just as hard taking it from handwritten work to computer text.

 

[MedPR]

I think this is good advice and would like to add that it's just as hard taking it from handwritten work to computer text.

Oh believe me I know. However, I came up with quite a few non-scientific explanations out of pure laziness and unwillingness to type out equations on here

 

[gettheleadout]

Oh believe me I know. However, I came up with quite a few non-scientific explanations out of pure laziness and unwillingness to type out equations on here

Were it not for pfaction's request for a step-by-step explanation I would have done the same haha...

I have never seen the pseudo-HH equation in this thread before, though it seems very useful. A disclaimer to pfaction is I would not want to use the method I did without a calculator on the MCAT. The shorter method is much more efficient. I'd rather someone else explain that method though.

 

[pfaction]

I'll review it tomorrow; but TBR used that equation and I'm still getting it wrong. near 2am so...

 

[kasho11]

I mean honestly using the HH equation isn't that much more work. It ends up just being

pKa = x^2 / [concentration given]

So you just multiply pKa by concentration and take the square root of it. That number is then the [H+] or [OH-] concentration, and take -log of that and you get pH or pOH.

Haha ok, written out it does seem like more work but it's easily doable in under a minute with scientific notation, even if you aren't adept at math.

 

[ipodtouch]

pKa = x^2 / [concentration given]

Just wondering, how do you get what you got above with the HH?
I understand how to solve the problem, but am having trouble making the correlation.

 

[pm1]

Your ICE table is set up correctly, but your calculation of Ka is off. We are given that pKa = 3.642, so since 3.624 = -log(Ka), Ka = 10^(-3.642) = 2.28E-4

How do you find Ka from pKa?
I was trying to the Kaplan method where you have the number multiplying by 10^-4 taken as a decimal subtracted from 4.
Ex: 2.28E-4 would be 4 - 0.228 = 3.78

Thank you!