Equilibrium of NH4F, (NH4)2S, and MgO in Water: Reactants or Products?

[ghostanime2001]

Homework Statement

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

The Attempt at a Solution

The only way i can determine if products or reactants are favoured is if i do a K_eq calculation. If K_eq > 1 then its products. If K_eq < 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 that's 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 I am stuck here in the very beginning of this question. PLEASE HELP.

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

 

[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.

 

[Borek]

So, the equilibria equation is actually $$NH_{4}^{+} + F^{-} \rightleftharpoons NH_{3} + HF$$

You can't ignore water.

--

 

[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?

 

[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^- K_a = 1.3 x 10^-13
K_b = 1 x 10^-7
K_b > K_a therefore HS^- will react as a base in solution

Li⁺ came from a strong base (LiOH) so Li⁺ will not hydrolyze but NO^2- will. NO^2- 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 NO^2- will be a base because it doesn't have another hydrogen to ionize. So I know NO^2- 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 ?

The answers:

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

Im in need of a lot of clarification.

 

[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.

 

[ghostanime2001]

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

 

[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.

 

[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,
$$[H_{3}PO_{4}][HPO_{4}^{2-}] = 10^{-14}$$

 

[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" don't have any ionizable hydrogen to behave as an acid. I am stuck with this one. Please help.

 

[queenofbabes]

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

 

[ghostanime2001]

What do you mean?

 

[queenofbabes]

When you combined your equations, your water disappears.

 

[ghostanime2001]

No it doesnt. This is what I get.

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

 

[queenofbabes]

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

 

[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.

 

[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:

 

[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.

 

[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.

 

[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.

 

[ghostanime2001]

Can you explain a little bit more clearer ?

 

[ghostanime2001]

Can you explain a little bit more clearer ? Also, how do you arrive at an answer such as in the worksheet with K_eq > 10²² ? how did they get such a big number and from where ?

 

[ghostanime2001]

You still haven't explained to me WHY and HOW in the worksheet there is that answer. You are not helping me when you just explain what YOU think. I want you to actually take a look in page 5 of those scanned pages i took the time to scan for you so you could help me out. If you still do not want to help me I'll take my post somewhere else where somebody else is better at explaining things than you guys. I was very patient with you BUT how much patience can I take..

 

[symbolipoint]

I tried to take a look at the scan #5, but after clicking once to open and once more to bring up a window with an enlarged version, neither of the two forms of the image were readable.

 

[ghostanime2001]

Make sure guys, that you download adobe reader *THE LATEST ONE* from adobe.com. If you still can't see it in ur browser window then download the picture and save it to wherever you want and then open it up with whatever default program ur computer uses to view images.


I also know that Mg(OH)2 is a somewhat strong base but not completely strong. Does it have a Kb ?

 

[queenofbabes]

It has a Kb, as all bases do regardless of their strength. A weak base simply has a smaller Kb.

For the time being I'm stuck with a pretty crap school laptop, the pics still don't work for me, neither does it offer me an option to download. Either way, homework helpers don't log on to this site 24/7, neither do they owe you anything, so pls keep your attitude in check.

 

[ghostanime2001]

go here:

http://ghostanime2.weebly.com/" <-- this link is my website that has the scans so you can see the method they used that I keep talking about. The method is located in a box under the category "G. Competitiion between Equilibria" in scan scan0004.jpg on the site.

 

[PhaseShifter]

(d) Hydrogen peroxide is a weak acid, while the hydrogen sulfite ion is a weak base.

hydrogen sulfide: $$H_{2}S$$

hydrogen sulfide ion: $$HS^{-}$$

hydrogen sulfite ion: $$HSO_{3}^{-}$$

Confusing these names can make a big difference when looking up constants for solubility and acid-base reactions.

 

[ghostanime2001]

I SOLVED All other problems except for the question MgO. I don't need help with other questions just that one (magnesium oxide) calculating either the reaction favours products or reactants USING my website scans method. ion There is a method to quantitatively calculate how much the reaction favours products or reactants. Don't tell me any other jargon from somewhere else. I want to know how to get that answer in the scanned worksheets off my site.

key = Look at post# 27 for directions to look specifically on the worksheets. This is the only way you'll see what I am talking about. PPl are talking about one thing and I'm talking about something else. And no wonder we can't solve this damn problem.

Would it hurt just to visit my site and look at the method they used and I'm using and also taking a look at the answer.

 

[PhaseShifter]

I can read the scans just fine, but there's no information given on how oxides react with water anywhere in the text. Unless they have an appendix with more Ka and Kb values, it seems like a very poorly written textbook in that it's asking you to to solve a problem without providing sufficient information.

They don't even drop a hint on any page that oxide is a much stronger base than hydroxide. (And even in a standard reference like the CRC Handbook of Physics and Chemistry you won't see a Kb for oxide, mainly because oxide is such a strong base you never see it in aqueous solution.)

I suspect they calculated the answer from thermodynamic tables instead, and didn't think it worthwhile to mention they were using a more advanced technique.

 

[ghostanime2001]

Why would they give me such a question that doesn't fit with the method they want me to use ? and that question seems fishy to me because there is no product acid only the reactant acid (water) unless they assumed something that makes the answer turn out to be inequality. And that assumption they think I should know which I really really really really don't know. I've tried and tried until my personality split and still cannot figure out how they came up with answer like that.

if there is anyone else on this forum that knows why the answer is an inequality then please help me with it. Your help is greatly appreciated but I am still not going to leave this question alone. And one thing I still can't figure out is how did they come up with answer as large as to the magnitude of 22 !

Nothing makes sense to me with that question.

 

[queenofbabes]

As mentioned before, this question is more than just acid-base equlibria. Take into account that Mg(OH)2 has a low Ksp as well, so consider its solubility equilibria.

I'm not sure how they got that number, but you can imagine that it'll be huge. Any OH- produced will precipitate out, lowering OH- concentration and driving the reaction forward to a virtual completion.

 

[ghostanime2001]

Do I just leave that question then?

 

[ghostanime2001]

Has anyone come across a question like this?