Calculating pH in an acid-base reaction

  • #1
QuantumCurt
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Homework Statement



Calculate the pH of a solution containing 5.0 g of [itex]NaHCo_3[/itex] (molar mass = 84.0 g) in 250. mL of H20. [itex]NaHCO_3[/itex] is amphiprotic. [itex]k_a= 4.7x10^{-11}[/itex], [itex]k_b=2.3x10^{-8}[/itex]

This is for a General Chemistry II class, for the record.




The Attempt at a Solution





I started by calculating the molarity of the NaHCO3 in the solution.

[tex] 5.0g x \frac{1 mole}{84.0 g} x \frac{1}{.250 L}=.238M[/tex]

With that calculated, I wrote out a balanced equation.
Then I set up an ICE (initial, change, equilibrium) chart

[tex] NaHCO_3 + H_2 O→N_aH_2CO_3^++OH^-[/tex]

With an initial molarity of .238 for NaHCO3, and values of 0 for NaCO3+ and OH-, I wrote an equilibrium constant expression for Ka

[tex] \frac{x^2}{.238-x}=4.7x10^{-11}[/tex]

Solving that for x, I get [itex]x=3.35x10^{-6}[/itex]

And taking the -log, I get a pH of 5.47.

The answer is supposed to be a pH of 9.9. What am I missing here? Is this something to do with the amphiprotic nature of NaHCO3? Am I writing an incorrect balanced equation?
 

Answers and Replies

  • #2
Qube
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Writing in the sodium ions and combining them to form things such as sodium carbonic acid is 1) confusing and 2) more than likely unrepresentative of the true chemistry in the system. Sodium hydrogen carbonate may actually exist in its unionized form in solution (if only in small quantities). On the other hand, I'm more than sure that sodium carbonic acid does not exist.

Therefore, if I were you, I would first write "the"* actual reaction. You might call this the net ionic reaction; I call this the correct reaction.

[itex]HOCO_{2}^{-} + H_{2}O \leftrightharpoons HO^{-} + H_{2}CO_{3}[/itex]

And yes, you are right, hydrogen carbonate ion is amphiprotic. So the question is now how will it behave in solution? As an acid, or as a base? Or as both?



The answer is both, of course, but the real question is what will the predominant behavior be?

*I say "the" because the reaction I wrote above is hardly the only equilibrium occurring in the system. Bicabonate ion is a wonderfully complex ion and it can react with water:

1) As an acid
2) As a base

And bicarb ion can also react with itself! This forms "carbonic acid" and carbonate ion.
 
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  • #3
QuantumCurt
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It's going to be behaving predominately as an acid since it's donating hydrogen ions, correct? Shouldn't that be [itex]H_2CO_3[/itex] though? The equation you've written has an uneven number of hydrogen ions. I was always taught that a balanced equation was necessary in these problems. Do I need to balance it with an additional 2H+?
 
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  • #4
QuantumCurt
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Wait...no...It's actually accepting protons, so it's acting as a base? I'm confused about these missing H+ ions.

edit-Never mind, I see you just edited it. That makes more sense now. It's acting as a base.
 
  • #5
Qube
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Shouldn't that be [itex]H_2CO_3[/itex] though? The equation you've written has an uneven number of hydrogen ions. I was always taught that a balanced equation was necessary in these problems.
You are correct! I wrote that reaction by mistake. I forgot about conjugate theory; the bicarb ion is behaving as a base in that reaction and so it becomes its conjugate acid, carbonic acid. Water in that reaction is the acid and becomes its conjugate base, the hydroxide ion.

It's going to be behaving predominately as an acid since it's donating hydrogen ions, correct?
Simply because something can donate hydrogen protons does not mean that it necessarily will to a large extent. Consider ammonia. We usually consider ammonia a base in water solution because even though it has three hydrogen protons - which can all be donated (at least in theory) - the formation of the amide ion, [itex]H_{2}N^{-}[/itex], is unfavorable (very much so in water solution). So try thinking about what K expressions stand for and what they mean.
 
  • #6
Qube
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It's acting as a base.
The bicarb ion sure is acting as a base in the equation I wrote above, but still, the following questions remain:

1) How do you know it is predominantly acting as a base?

2) If you have another amphiprotic substance, how do you know whether it is a base or an acid if I didn't write out the main equilibrium?
 
  • #7
QuantumCurt
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Got it!!

I had been thinking that it was behaving as an acid, so I was setting my equilibrium constant expression equal to Ka. I set it equal to Kb this time to get my OH- concentration, took the -log and got a pOH of 4.13. Then I subtracted that from 14 and got the correct pH of 9.9.

I haven't seen these for a couple months. The nuances of them slipped my mind. I'm ready for these finals to be over with.

Thanks a lot for the help!!
 
  • #8
QuantumCurt
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The bicarb ion sure is acting as a base in the equation I wrote above, but still, the following questions remain:

1) How do you know it is predominantly acting as a base?

2) If you have another amphiprotic substance, how do you know whether it is a base or an acid if I didn't write out the main equilibrium?
This is a topic that I've always been confused about in these types of problems. When dealing with amphiprotic substances, I've never really understood the rules for determining what it behaves as.
 
  • #9
Qube
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This is a topic that I've always been confused about in these types of problems. When dealing with amphiprotic substances, I've never really understood the rules for determining what it behaves as.
Try writing out the Ka and Kb expressions for hydrogen carbonate ion. Let them talk to you.

Got it!!

I had been thinking that it was behaving as an acid, so I was setting my equilibrium constant expression equal to Ka. I set it equal to Kb this time to get my OH- concentration, took the -log and got a pOH of 4.13. Then I subtracted that from 14 and got the correct pH of 9.9.

I haven't seen these for a couple months. The nuances of them slipped my mind. I'm ready for these finals to be over with.

Thanks a lot for the help!!
Great! Now what if your test had these choices:

a) pH = 9.9
b) pH = 5.5

So, we're about to untangle this. Just let the K expressions start talking to you. Remember that in general:

[tex]K = \frac{[product]}{[reactant]}[/tex]

So think about the implications of that fraction and the value that the fraction is set equal to.
 
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  • #10
QuantumCurt
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Can I determine that it's acting as a base simply because I end up with OH- as one of the products? If it was acting as an acid, it would have gone to H30+ as one of the products, right?
 
  • #11
QuantumCurt
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Is it because HCO3- is a negative species interacting with a neutral species (H2O)?
 
  • #12
Qube
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Can I determine that it's acting as a base simply because I end up with OH- as one of the products? If it was acting as an acid, it would have gone to H30+ as one of the products, right?
The problem with this is that there are multiple, simultaneous equilibria in the system. Here are 4. As we can see, we end up with both hydroxide ion and hydronium ion as products. The question now is 1) which one matters and 2) to what extent does the one that matters reach? (Assuming there is only equilibrium that matters, but as far as I can tell, your general chemistry class has not gone very much into depth with consideration of acid/base systems).


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Is it because HCO3- is a negative species interacting with a neutral species (H2O)?
Negatively charged species can also be fairly strong acids - consider [itex]HOSO_{3}^{-}[/itex], which has a Ka of 1.2 * 10-2. Not a strong acid in water per se but definitely up there on the totem pole for weak acids.

So again, we must go back to our K expression. Write out the equations corresponding to the reactions of hydrogen carbonate ion as an acid and as a base and write out their corresponding K expressions. Set these K expressions equal to the appropriate values. Let this all talk to you. For example, if we have a large K value, what does this say about the ratio of products to reactants? What about small? What are the relative sizes of Ka and Kb for hydrogen carbonate ion? What may you conclude from this?
 
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  • #13
QuantumCurt
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If the concentration of H+ is greater than the concentration of OH-, it's a acidic...and if the concentration of H+ is less than OH-, it's basic. If the two are equal, then it's neutral?

That's what I'm seeing from looking back in my notes. I don't recall us ever even really going too in depth about how to determine that.
 
  • #14
Qube
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That is right but how do you know the concentration of either a priori? Again, please consider the K expressions in the ways I mentioned above.
 
  • #15
QuantumCurt
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The larger the Ka, the more acidic something is. The larger the Kb, the more basic it is, correct?
 
  • #16
Qube
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The larger the Ka, the more acidic something is. The larger the Kb, the more basic it is, correct?

Sure. What about for the same compound?
 
  • #17
QuantumCurt
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It would be neutral then, wouldn't it?
 
  • #18
Qube
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It would be neutral then, wouldn't it?

Hydrogen carbonate ion has both a significant Ka and a Kb. I think we can agree that a solution of it is not neutral.
 
  • #19
QuantumCurt
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Okay, then I'm really not seeing what these different expressions tell me. All I'm seeing is two identical expressions that are set equal to two different values.
 
  • #20
Qube
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Okay, then I'm really not seeing what these different expressions tell me. All I'm seeing is two identical expressions that are set equal to two different values.

Write out your K expressions here or otherwise show me (i.e. link me to a picture of your work.) Your K expressions should differ by more than just what they're equal to.
 
  • #21
QuantumCurt
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Okay, then I'm really not seeing what these different expressions are telling me. All I'm seeing are two identical expressions that are set equal to different values.
 
  • #22
QuantumCurt
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Whoops, double post. It wasn't showing up so I posted again
 
  • #23
Qube
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Something suggests to me that you are not writing your K expressions correctly. Seeing the K expressions for what they mean is very important in chemistry rather than just accepting a "rule."
 
  • #24
QuantumCurt
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I'll get back to it later. I have this final on Monday, and I still have a ton of stuff to go over. I can't spend too much time on this one thing.

When I've got time later I'll take another look at it and post it here.

Thanks a lot for the help!
 
  • #25
Qube
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Good luck! Hopefully some time off will lend you more insight onto what K expressions mean.
 

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