# Acid/Base Reactions

1. Apr 12, 2005

### eku_girl83

1) State whether aqueous solutions of the following salts are acidic, basic, or neutral.

a) NH4N03
I think this is acidic due to the hydrolysis of the NH4 ion, which produces H+ ions?

b) KH2PO4
not sure about this one, since it is a diprotic acid and ionizes in more than one step
any hints?

c) NaBr
this is a strong electrolyte, so it disassociates completely in water; hence, common ion effect is not observed

Are these correct? Can someone please give me a hint on part b?

2. Apr 12, 2005

### Staff: Mentor

a. OK
b. You are right - this one is tricky. Solution is slightly acidic. The last dissociation step can be neglected. First two are competing - hydrolysis of first against dissociation of teh second. I know the result (using BATE) but am a little bit to tired at the moment to see how to do the calculations properly. Besides, I made BATE just to be not forced to do such calculations by hand
c. No idea what common ion effect has to do, but pH will be neutral. Or a little bit less than neutral, as NaOH is not as strong base as everybody thinks.

Last edited by a moderator: Aug 13, 2013
3. Apr 12, 2005

### so-crates

For (b), Since the pKa of H2PO4- is 6.8 which is less than 7, the solution will be slightly acidic. This is all we have to worry about since we don't have any other salts or acids with common ions (if you added H3PO4 for instance, the solution would become a more acidic buffer)

4. Apr 13, 2005

### GCT

b may be basic, it is the conjugate of a weak acid, thus it may be a significant base (it's somewhat similar to calculating the equivalence point pH of a weak acid/strong base titration). It all depends on whether it will have more free energy in basic reactions or as an acid, and no I don't believe that both will occur, one will predominate.

5. Apr 13, 2005

### Staff: Mentor

b is not basic

concentration of sodium dihydrate phosphate vs pH of the solution:

1M -> 4.68
0.1M -> 4.69
0.01M -> 4.79
0.001M -> 5.13
0.0001M -> 5.61
0.00001M -> 6.11

Last edited by a moderator: Aug 13, 2013
6. Apr 13, 2005

### GCT

You simply calculated the pH as if KH2P04 were not an anion, there is no common ion effect preventing it from hydrolyzing water, you did not take this into account; it can react both as an acid and a base...similar to water (except that water reacts with itself).

You'll need to see if the hydrolysis reaction predominates over, dissociation reaction.

7. Apr 13, 2005

### Staff: Mentor

How do you know what I did? Have you calculated anything on your own?

Results presented are for the full calculations of all ions and all equilibrium present in the solution.

For example for the 0.1M concetration:

[H+] = 2.049e-5 (pH = 4.69)
[OH-] = 4.881e-10
[H3PO4] = 2.864e-4
[H2PO4(-)] = 0.09941
[HPO4(2-)] = 3.068e-4
[PO4(3-)] = 6.690e-12
[KOH] = 1.543e-10
[K+] = 0.1

The only way you can prove I am wrong and I have not did full calculations is to show which of the equilibrium equations is not satisfied. You either find the equilibrium not satisfied or you have lost.

For your information - constants used:
pKw = 14
pKa1 = 2.15
pKa2 = 7.20
pKa3 = 12.35
pKb1 = 0.50 (yes, that's pKb for KOH, I can quote source of this value)

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8. Apr 13, 2005

### GCT

Obviously my message is not getting across. I'm not completely assuming that you're incorrect however that you've neglected the possibility that KH2P04 can act as a base rather than an acid. I don't need to work anything out to see that you've neglected its possible role as a base...every one of your pH is acidic. The pH will be completely p basic or acidic.

I don't have the time to do the calculations myself at the moment, might get back to it later.

again, this is for the assumption that it will react completely as an acid. However, note the situation if the hydrolysis reaction to predominate...La Chatelier's principles indicates that the reaction $$H2PO_4^{-1} \leftrightharpoons HPO4^{2-}+H_3O$$ will go to the left to refurnish the reactant. So there is no acidic contribution in the end.

Well, I'm not really trying to prove you wrong, if the dissociation reaction predominates...you're probably right. I admit that even my proposal may be wrong, however you don't seem to be understanding my point. Again, I'm saying that the reaction has two possibilities, yours may be right....or wrong, got it?

nobody's winning or losing here

9. Apr 13, 2005

### so-crates

Lets consider a related question

What will be the pH of a 1M H3PO4 solution? The pKa of H3PO4 is 2.

What will be the pH of a 1 M H2PO4- solution? The pKa of H2PO4- is 6.8.

I originally though this problem was harder than I thought it was. Then I realized the ONLY thing that matters is the pKa of H2PO4-. Yes, H2PO4- can act as a base. So can H3PO4, theoretically. It doesn't matter. Since we know the pKa, we know whether the solution will be acidic or basic. If you added say HPO4 2-, then it would be more comliplicated, but its not.

Edit: I got those pKa numbers off the web so I don't know if they are correct or not. If the pKa given is < 7, its acidic, if > &, then it will be basic.

Last edited: Apr 13, 2005
10. Apr 13, 2005

### Staff: Mentor

Seems you don't understand very basic thing - there is no such thing like neglecting KH2P04 reacting like base or acid if calculations take into account both equiliribria H2PO4(-) is involved in (described by Ka1 and Ka2 constants).

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11. Apr 13, 2005

### Staff: Mentor

1.09

4.68

But I have used slightly different constants - 2.148 and 7.199 (pKa3 is 12.35 to my best knowledge).

Check BATE to play with such calculations.

Last edited by a moderator: Aug 13, 2013
12. Apr 14, 2005

### GCT

common method used to create buffer solutions

$$H2PO4^{-}_{aq}+H2O_{l} \rightleftharpoons OH^{-} +H3PO4_{aq}$$

Let me put it very slowly for you. Consider the first Ka, the first dissociation constant, which both of you have completely neglected.

since you guys are so adamant about Ka...

-what does Ka pertain to...equilibrium concentrations
$$Ka= \frac{[H2PO4^{-}][H3O+]}{[H3PO4]}$$

what do you imagine would happen when we have an initial H2PO4- concentration and no H3PO4, it will refurnish the equilibrium concentrations so that the concentrations in the equation mathematically satisfy Ka.

Actually this method a common way in which buffers are made in the lab, by simply dissolving a certain about of the pure conjugate base, and at equilibrium you'll have a buffer system of the conjugate base with its acid.

Unfortunately I don't believe that the henderson-hasselbach equation will accurately describe this buffer solution. It is derived from the Ka equation, one will need to calculate the similar equation with Kb

Socrates, read up on Ka and pKa it is mathematically specific, not a prophylatic equation to be used for every situation.

13. Apr 14, 2005

### GCT

anyone who has taken second year gen chem lab should know this method of creating buffer solutions

14. Apr 14, 2005

### Staff: Mentor

Regardless of what you are writing Ka1 is taken into account.

Show me which of the equlibrium equations is not satisfied.

Last edited by a moderator: Aug 13, 2013
15. Apr 14, 2005

### GCT

simply put, part b on the original post refers to a common method of preparing buffer solutions, I should have recognized it from the start.

Your equations are not satisfied in that the whole basis of your calculations is incorrect. I'm not saying there's anything wrong with your BATE, just your approach to this problem. You should have focused on the first Ka (and Kb) and have noticed specifically that the base is the only agent that will establish the equilibrium and satisfy Ka (and Kb) mathematically.

16. Apr 14, 2005

### Staff: Mentor

Please stop questioning right answers as you create confusion and mislead posters. You simply don't understand the most basic stuff of chemical equilibrium.

EOT

Last edited by a moderator: Aug 13, 2013
17. Apr 14, 2005

### Gokul43201

Staff Emeritus
Let's try and keep this civil, wot ?

I did a very crude calculation for 1M KH2PO4 and got a number that is a little less than pH = 5, but know for certain that pH < 6.0 (at 1.0M)

18. Apr 14, 2005

### GCT

How did you obtain a acidic pH? You all need to read up on buffer preparation.

$$Kb= \frac{[OH-][H3PO4]}{[H2PO4-]}$$

$$Kb= \frac{[x][x]}{[[H2PO4-]-x]}$$

solve for $[OH-]=[x]$

also, relating to my previous post

$$pOH=pKb+log([H3PO4]/[H2PO4-])$$

See how $Ka2$ compares with $Kb1$ in value

H2PO4- is amphoteric in can react as an acid or a base, it is a base in terms of the first equilibrium, an acid in terms of the second equilibrium.

phosphate buffer is commonly prepared by dissolving a certain amount of pure H2PO4- and adding a strong acid to adjust the buffer pH. Thus it is a base.

19. Apr 14, 2005

### GCT

If I'm doing anything wrong specifically somebody let me know.

Just take the case where a acid/base titration is performed, weak acid with a strong base. At equivalence point, where all of the weak acid has been consumed you'll be left with the salt. For instance

$$CH3COOH+NaOH \leftrightharpoons~NaCH3COO-+H20$$

The salt is NaCH3COO-, the CH3COO-, being the conjugate of a weak acid, will be a significant base.

$$CH3COO-+H2O \rightleftharpoons~CH3COOH+OH-$$

thus the solution will be basic at equivalence point

similar case with part b of the OP's post

20. Apr 15, 2005

### GCT

I'm asking a professor about this situation, waiting for an answer at the moment. Either of us can be right about this situation...despite this, I'll try clearing up everything in the end.

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