# Products or Reactants ?

1. Aug 16, 2009

### ghostanime2001

1. The problem statement, all variables and given/known data
When the following substances are introduced into water do the resulting equilibria favour products or reactants?

a) NH4F

These are other questions like this one:
b)(NH4)2S
c)MgO

3. The attempt at a solution

The only way i can determine if products or reactants are favoured is if i do a Keq calculation. If Keq > 1 then its products. If Keq < 1 then its reactants.

$$NH_{4}F \rightleftharpoons NH4^{+} + F^{-}$$

This is again a hydrolysis reaction because i have to determine which ion is the spectator ion. It says in my book that:
"Spectator ions can be thought of as being the result of dissolving a strong base or a strong acid. The positive ion (cation) is considered to result from a base, BOH, and the negative ion (anion) is considered to result from an acid, HA"

If thats correct then the cation (NH4+) came from NH3 and the anion (F-) came from HF so that the entire equilibria is:

$$NH_{3} + HF \rightleftharpoons NH_{4}F + H_{2}O$$

But neither NH3 or HF are strong acids or bases. I'm stuck. How can I proceed through with the rest of this question ? I still don't understand Hydrolysis that im stuck here in the very beginning of this question. PLEASE HELP.

The answer btw is:
reactants; Keq = 8.5 x 10-7

2. Aug 16, 2009

### Fightfish

Firstly, $$NH_{4}F \rightleftharpoons NH4^{+} + F^{-}$$ is wrong. An ionic salt dissociates completely in aqueous media.
Construct the individual hydrolysis equations for the ions:

$$NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{3} + H_{3}O^{+}$$

$$F^{-} + H_{2}O \rightleftharpoons HF + OH^{-}$$

So, the equilibria equation is actually $$NH_{4}^{+} + F^{-} \rightleftharpoons NH_{3} + HF$$, which yields
$$K_{C} = \frac{[NH_{3}][HF]}{[NH_{4}^{+}][F^{-}]}$$
From the Ka and Kb values of the hydrolysis of these ions, the ratios can be computed.

3. Aug 16, 2009

### Staff: Mentor

You can't ignore water.

--

4. Aug 16, 2009

### ghostanime2001

both ions come from weak acid/base is that the reason why you included the hydrolysis of both ions ?

WoW... ur right i got 8.787878 x 10-7 is it close enough to the answer in a couple post prior?

5. Aug 16, 2009

### ghostanime2001

I have two other problems which I COMPLETELY DO NOT UNDERSTAND. I'm serious.

c) H3PO4 and Na2HPO4
d) H2O2 and kHS
e) MgO
f) H2S and LiNO2

starting with c):

I separated H3PO4 and Na2HPO4 into ions:
$$H_{3}PO_{4} \rightarrow H_{2}PO_{4}^{-} + H^{+}$$
$$Na_{2}HPO_{4} \rightarrow 2Na^{+} + HPO_{4}^{2-}$$

and combining the anions:
$$H_{2}PO_{4}^{-} + HPO_{4}^{2-} \rightleftharpoons HPO_{4}^{2-} + H_{2}PO_{4}^{-}$$

$$Keq = \frac{K_{a} \mbox{(reactant acid)}}{K_{a} \mbox{(product acid)}} = \frac{ \mbox{6.3 x 10^{-8}}}{ \mbox{6.3 x 10^{-8}}}$$

There is something wrong with that, what am I doing wrong ?

starting with d):

$$H_{2}O_{2} \rightarrow HO_{2}^{-} + H^{+}$$
$$KHS \rightarrow K^{+} + HS^{-}$$

$$HO_{2}^{-} + HS^{-} \rightleftharpoons O_{2}^{2-} + H_{2}S$$
or
$$HO_{2}^{-} + HS^{-} \rightleftharpoons H_{2}O_{2} + S^{2-}$$

This is so confusing

starting with e):

$$MgO \rightarrow Mg^{2+} + O^{2-}$$

$$Mg^{2+} + 2H_{2}O \rightleftharpoons Mg(OH)_{2} + 2H^{+}$$

$$O^{2-} + 2H_{2}O \rightleftharpoons H_{2}O + 2OH^{-}$$
-------------------------------------- (You told me not to ignore water so here it is)
$$Mg^{2+} + O^{2-} + 4H_{2}O \rightleftharpoons Mg(OH)_{2} + H_{2}O + 2H_{2}O$$

Simplifies down to:
$$Mg^{2+} + O^{2-} + H_{2}O \rightleftharpoons Mg(OH)_{2}$$

Now, HOW can you tell which one are acids/bases.

Starting with f):

$$H_{2}S \rightarrow HS^{-} + H^{+}$$
$$LiNO_{2} \rightarrow Li^{+} + NO^{2-}$$

HS- Ka = 1.3 x 10-13
Kb = 1 x 10-7
Kb > Ka therefore HS- will react as a base in solution

Li+ came from a strong base (LiOH) so Li+ will not hydrolyze but NO2- will. NO2- will act as a base in solution.

BUT HOW CAN THEY BOTH BE BASES? One of them has to be an acid and one of them a base. I know that NO2- will be a base because it doesnt have another hydrogen to ionize. So I know NO2- is going to be a base and HS- will be an acid when the anions are mixed.

$$HS^{-} + NO^{2-} \rightleftharpoons HNO_{2} + S^{-2}$$

$$K_{eq} = \frac{K_{a} \mbox{(reactant acid)}}{K_{a} \mbox{(product acid)}}= \frac{ \mbox{1.3 x 10^{-13}}}{ \mbox{5.1 x 10^{-4}}}$$

$$= 2.5 x 10^{-10}$$ (which is WRONG!)

What can I do now ?

c: products; Keq = 1.1 x 105
d: reactants; Keq = 2.4 x 10-5
e: products; Keq > 10^22 (acid = H2O)
f: reactants; Keq = 2.0 x 10-4

Im in need of a lot of clarification.

Last edited: Aug 16, 2009
6. Aug 17, 2009

### Fightfish

Sorry don't have much time atm, so will only answer c) for now. Will look at rest later.
The equilibria in (c) should be individually:
$$H_{3}PO_{4} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + H_{3}O^{+}$$
$$HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + OH^{-}$$

Combining:
$$H_{3}PO_{4} + HPO_{4}^{2-} \rightleftharpoons 2H_{2}PO_{4}^{-}$$,
(Phosphoric acid is a weak acid and only dissociates partially.)
which yields
$$K_{eq} = \frac{[H_{2}PO_{4}^{-}]^{2}}{[H_{3}PO_{4}][HPO_{4}^{2-}]} = \frac{K_{a}(H_{3}PO_{4}) K_{b}(HPO_{4}^{2-})}{10^{-14}}$$

Qualitatively, this is an acid-base reaction, and thus the position of the equilibrium will lie on the side of the products, as the depletion of the H+ and OH- ions drives the individual equilibria towards the formation of products.

Last edited: Aug 17, 2009
7. Aug 17, 2009

### ghostanime2001

How did you get 10-14 for the denominator ? and why? I know its Ka for H2O but why ?

8. Aug 17, 2009

### ghostanime2001

and also what method are you using? The method in my worksheet is kinda different. I have scanned the sheet for you so you can see that different method.

9. Aug 17, 2009

### queenofbabes

$$[H_{3}O^{+}][OH^{-}] = 10^{-14}$$
But from your equations $$[H_{3}PO_{4}] = [H_{3}O^{+}]$$ and $$[HPO_{4}^{2-}] = [OH^{-}]$$, not forgetting you started with only the reactants. Thus,
$$[H_{3}PO_{4}][HPO_{4}^{2-}] = 10^{-14}$$

10. Aug 18, 2009

### ghostanime2001

I figured out all the other questions. I did another way and got the same answer. The only one giving me trouble is e) MgO

To see my method:

$$H_{3}PO_{4} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{-} + H_{3}O^{+}$$

$$Na_{2}HPO_{4} \rightarrow 2Na^{+} + HPO_{4}^{2-}$$

$$HPO_{4}^{2-} + H_{2}O \rightleftharpoons H_{2}PO_{4}^{2-} + OH^{-}$$

Combined:
$$H_{3}PO_{4} + HPO_{4}^{2-} \rightleftharpoons H_{2}PO_{4}^{-} + H_{2}PO_{4}^{-}$$ (Note that i didnt include $$2H_{2}PO_{4}^{-}$$ just so I could see the distinction between which species reacted as acids/bases)

$$K_{eq} = \frac{K_{a} \mbox{(product acid)}}{K_{a} \mbox{(reactant acid)}} = \frac{7.1 \mbox{x} 10^{-3}}{6.3 \mbox{x} 10^{-8}} = \mbox{112698.4127 which is approx. 1.1 x 10^{5}}$$

Now, for that troublesome e) MgO

I know that its a salt and in my worksheet it said all salts are soluble and 100% ionized so,

$$MgO \rightleftharpoons Mg^{2+} + O^{2-}$$

$$Mg^{2+} + 4H_{2}O \rightleftharpoons \underbrace{Mg(OH)_{2}}_{ \mbox{That is my first guess}} + 2H_{3}O^{+}$$

$$O^{2-} + H_{2}O \rightleftharpoons OH^{-} + OH^{-}$$

When combined:
$$Mg^{2+} + O^{2-} \rightleftharpoons Mg(OH)_{2} + OH^{-}$$

Now wheres $$K_{a} \mbox{(reactant acid)}$$ the "reactants" dont have any ionizable hydrogen to behave as an acid. I am stuck with this one. Please help.

11. Aug 18, 2009

### queenofbabes

You forgot the water again, methinks. (look at the equations you stated more carefully) H2O is the acid in this case.

Last edited: Aug 18, 2009
12. Aug 18, 2009

### ghostanime2001

What do you mean?

13. Aug 18, 2009

### queenofbabes

When you combined your equations, your water disappears.

14. Aug 18, 2009

### ghostanime2001

No it doesnt. This is what I get.

$$Mg^{2+} + O^{2-} + H_{2}O \rightleftharpoons Mg(OH)_{2}$$

Last edited: Aug 18, 2009
15. Aug 18, 2009

### queenofbabes

Exactly. You had left out the H20 in post #10. So in this case H2O is your "acid".

16. Aug 18, 2009

### Fightfish

The key equilibria should be:
$$O^{2-} + H_{2}O \rightleftharpoons 2OH^{-}$$ (almost a direct arrow!)
$$Mg^{2+} + 2OH^{-} \rightleftharpoons Mg(OH)_{2}$$
The second one is essentially ignorable due to the relatively low Ksp of magnesium hydroxide, and given that we expect a very very high concentration of hydroxide ions.

17. Aug 18, 2009

### ghostanime2001

what about $$K_{a} \mbox{(product)}$$ in post #14 I know now thanks to you that $$K_{a} \mbox{(reactant)}$$ is $$H_{2}O$$ in post #14

All these problems came from these worksheets:

Last edited: Aug 18, 2009
18. Aug 18, 2009

### ghostanime2001

Two last sheets giving a total of 5 pages.

I just hope someone can explain these problems to me so i can understand them faster.

19. Aug 19, 2009

### Fightfish

There's no $$K_{a} \mbox{(product)}$$ for the second equilibria, which is not a dissociation of an acid or a base, but a solubility equilibria. So, you can treat the second equilibria as negligible as the high concentration of OH- from equilibria one would drive the second equilibria to the extent that most of the Mg2+ would be precipitated out as the product.

20. Aug 19, 2009

### Fightfish

(d) Hydrogen peroxide is a weak acid, while the hydrogen sulfite ion is a weak base.
Thus,
$$H_{2}O_{2} + HS^{-} \rightleftharpoons HO_{2}^{-} + H_{2}S$$

(f) Once again, hydrogen sulfite is a weak acid, and thus you do not dissociate it and work with the conjugate base. Hence,
$$H_{2}S + NO^{2-} \rightleftharpoons HS^{-} + HNO_{2}$$

Key takeaway: Leave the weak acids and bases as they are and don't dissociate them completely to give the conjugate bases/acids and construct your equilibria from there -> the equilibria starts with the weak acids/bases, which only partially dissociate in water.