TBR chem Determining pH for Weak Reagents shortcut

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Steelersfan2009

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Hi, TBR gave this formula for determining the pH of weak reagents. I'm just having trouble understanding it. What I'm having trouble with is it telling me when I can or cannot use this equation.

pH = 1/2 pKa - 1/2 log [HA]

Here is what the book says. According to equation increasing [HA] lowers the pH. Stronger acids have more dissociation, so they should form solutions of a lower pH.

So they use this to find pH of a weak acid correct?

Here is a example problem

What is the pH of 1.00 M HF with pKa = 3.32?
answer: 1.66

Here is the solution.
*** it gave 2 requirements that I'm having trouble understanding
1) the weak acid concentration must be greater than Ka
2) pKa must lie between 2 and 12.
BOTH of these creteria are met??

Can someone explain to me how these 2 criteria are met.. I'm so horrible at this part of Acid and base :mad:

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Hi, TBR gave this formula for determining the pH of weak reagents. I'm just having trouble understanding it. What I'm having trouble with is it telling me when I can or cannot use this equation.

pH = 1/2 pKa - 1/2 log [HA]

Here is what the book says. According to equation increasing [HA] lowers the pH. Stronger acids have more dissociation, so they should form solutions of a lower pH.

So they use this to find pH of a weak acid correct?

Here is a example problem

What is the pH of 1.00 M HF with pKa = 3.32?
answer: 1.66

Here is the solution.
*** it gave 2 requirements that I'm having trouble understanding
1) the weak acid concentration must be greater than Ka
2) pKa must lie between 2 and 12.
BOTH of these creteria are met??

Can someone explain to me how these 2 criteria are met.. I'm so horrible at this part of Acid and base :mad:

Yes, both of these criteria are met:

1) You're told it's a weak acid with a 1.00 M concentration-the PKA is 3.32. pKa=negative log of the Ka. If I told you the ka was around 1 x 10^-3, then the pka would be around 3, right? Surely, 1 x 10^-3 is less than 1.00 M. Thus, the concentration is greater than the Ka. For this equation, the concentration is usually greater than the Ka (since we're dealing with weak acids, which have small Kas)

2) the pka is 3.32, which is between 2 and 12.
 
so if this is the shortcut, what is the "usual" way to solve it?
It's not the H-H equation is it? with just that information?

I'm wondering how i would solve this not knowing that TBR shortcut (I've never seen it before)

thanks!
 
so if this is the shortcut, what is the "usual" way to solve it?
It's not the H-H equation is it? with just that information?

I'm wondering how i would solve this not knowing that TBR shortcut (I've never seen it before)

thanks!

You would need to do ICE tables to find the [H+], and then use pH=-log[H+].

And no, H-H is for buffers and conjugate acid/base pairs.
 
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uh-oh, this is not good. Whats an ICE table?

Forgot equilibrium? Say we have a weak acid, HA.

HA--> H+ + A- (of course we have equilibrium arrows)

You'd be given the Ka, the initial concentration of the weak acid (say its 1.00 M), and would be asked the pH of the weak acid.

HA --> H+ + A-
I 1.M 0 0

C -x +x +x

E (1.00-x) +x +x

Ka=products/reactants, (x)(x)/1.00 (ignore the x in 1.00-x)

x^2=Ka(1.00)M, x=sqrt of Ka. Once you solve for x, which is the H+ concentration, use pH=-log[H+], and voila, you have the pH. Oh, and I=initial, C=change, E=equilibrium (the table looks messed up, it won't let me do it the way I want). The reason you do ICE tables is because a weak acid does NOT fully dissociate, so you don't know what the H+ concentration is to use pH=-log [H+]. You need to find out just how much dissociates in order to plug it in that equation, whereas if that were a SA you could've just used the given concentration because you know a SA fully dissociates. Hope that helps.
 
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so if this is the shortcut, what is the "usual" way to solve it?
It's not the H-H equation is it? with just that information?

I'm wondering how i would solve this not knowing that TBR shortcut (I've never seen it before)

thanks!

It's painful. You'd have to set up an ICE table to determine how many protons dissociate into solution based on the Ka of HF.

Your equation would look something like

Ka = x^2/(x-1)

Since this is a weak acid we can ignore the x in the denominator, so the denominator is essentially 1.

Now

Ka = x^2 (you would need to compute the Ka based on the pKa of the acid as well)

x = sqrt(4.79e-4) =2.187e-2

It turns out, based on the ICE table that x is the concentration of protons in solution.

ph = -log(x) = -log(2.187e-2) = 1.66

So this way also works, which is the way I was taught, but the equation they give you is much much faster.
 
Forgot equilibrium? Say we have a weak acid, HA.

HA--> H+ + A- (of course we have equilibrium arrows)

You'd be given the Ka, the initial concentration of the weak acid (say its 1.00 M), and would be asked the pH of the weak acid.

HA --> H+ + A-
I 1.M 0 0

C -x +x +x

E (1.00-x) +x +x

Ka=products/reactants, (x)(x)/1.00 (ignore the x in 1.00-x)

x^2=Ka(1.00)M, x=sqrt of Ka. Once you solve for x, which is the H+ concentration, use pH=-log[H+], and voila, you have the pH. Oh, and I=initial, C=change, E=equilibrium (the table looks messed up, it won't let me do it the way I want). The reason you do ICE tables is because a weak acid does NOT fully dissociate, so you don't know what the H+ concentration is to use pH=-log [H+]. You need to find out just how much dissociates in order to plug it in that equation, whereas if that were a SA you could've just used the given concentration because you know a SA fully dissociates. Hope that helps.


HAHAHA.....thanks for putting up with me. I know that stuff! I just don't have the fancy ICE name for it. I'm just a simple guy trying to get a decent MCAT score....nothing frilly over here.....but thanks!

you freaked me for a second......
 
It's painful. You'd have to set up an ICE table to determine how many protons dissociate into solution based on the Ka of HF.

Your equation would look something like

Ka = x^2/(x-1)

Since this is a weak acid we can ignore the x in the denominator, so the denominator is essentially 1.

Now

Ka = x^2 (you would need to compute the Ka based on the pKa of the acid as well)

x = sqrt(4.79e-4) =2.187e-2

It turns out, based on the ICE table that x is the concentration of protons in solution.

ph = -log(x) = -log(2.187e-2) = 1.66

So this way also works, which is the way I was taught, but the equation they give you is much much faster.

yes that is a kick ass shortcut indeed......and makes me regret not using BR more and more by the day! TPR is hardcore about math though so that kinda stuff doesn't bother me.......of course once i time myself it could be a different story

cheers
 
Yes, both of these criteria are met:

1) You're told it's a weak acid with a 1.00 M concentration-the PKA is 3.32. pKa=negative log of the Ka. If I told you the ka was around 1 x 10^-3, then the pka would be around 3, right? Surely, 1 x 10^-3 is less than 1.00 M. Thus, the concentration is greater than the Ka. For this equation, the concentration is usually greater than the Ka (since we're dealing with weak acids, which have small Kas)

2) the pka is 3.32, which is between 2 and 12.


btw thanks for the clarification, I see that it has already helped out others!
 
Ka = x^2 (you would need to compute the Ka based on the pKa of the acid as well)

x = sqrt(4.79e-4) =2.187e-2

It turns out, based on the ICE table that x is the concentration of protons in solution.

ph = -log(x) = -log(2.187e-2) = 1.66

So this way also works, which is the way I was taught, but the equation they give you is much much faster.

The above calculations can actually be done much more easily. Don't convert the pKa to what you have up there.
Just do this:

pKa=-log(Ka)

Ka=10e-pKa = 10e-3.32

Now, since

x=sqrt(Ka)

so X = sqrt(10e-3.32) = 10e-1.66

So pH=-log(10e-1.66) = 1.66

Voila!

Notice you don't even need a calculator for this.

If you do the log math you'll notice that that is essentially the TBR trick.
 
Hi, TBR gave this formula for determining the pH of weak reagents. I'm just having trouble understanding it. What I'm having trouble with is it telling me when I can or cannot use this equation.

pH = 1/2 pKa - 1/2 log [HA]

Here is what the book says. According to equation increasing [HA] lowers the pH. Stronger acids have more dissociation, so they should form solutions of a lower pH.

So they use this to find pH of a weak acid correct?

Here is a example problem

What is the pH of 1.00 M HF with pKa = 3.32?
answer: 1.66

Here is the solution.
*** it gave 2 requirements that I'm having trouble understanding
1) the weak acid concentration must be greater than Ka
2) pKa must lie between 2 and 12.
BOTH of these creteria are met??

Can someone explain to me how these 2 criteria are met.. I'm so horrible at this part of Acid and base :mad:
Thanks for sharing! I found this helpful!
 
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