Question

Hello Members of OpenStudy.

I am taking a minor course in Chemistry for the next Semester, so I have to catch up on a few things. I will post my question and my reasoning in the comment below, showing complete work. I would appreciate if someone could check it with me and tell me if I am on the right track and if my reasoning is correct from a chemical point of view.

Answer 1

Medical Topic: Adamantine, dental enamel.

\[\large Ca_5(PO_4)_3OH + 2 H_20 \longrightarrow 5Ca^{2+}+3PO_4^{3-}+OH^{-} \]

Considering this Reaction Equation, Adamantine reacts with saliva (mainly Water for this purpose).

For this I have to check how soft drinks that have small bits of an acid character distort the balance of the above equation.

Now my reasoning:
If the soft drink has a slight acid Character, you drink it and it will react with the Water \(H_2O\) to form \(H_3O^+\). On the side of the Educts there are Hydroxide Ions present.

So the balance will be distorted, \( H_3O^+ + OH^- \longrightarrow H_2O\) which yields to an increase of water in the side of the Products.

Considering the Principle of Le Chatelier, the distortion will cause the reaction to choose the side of products, causing an increase in the reduction of Adamantine.

Answer 2

\[H_3O^+ + OH^- \longrightarrow H_2O \] **Correction, which leads to an increase of water in the side of the \(\underline{EDUCTS}\)

Answer 3

First, the compound on the left is not reduced. The oxidation numbers are +2 (Ca), +5 (P), -2 (O) and +1 (H) on both left and right, so this is not a redox reaction.

Second, the way to solve this is to understand that there is another equilibrium going on in water solution, always, and that is the autoionization reaction of water with itself:

2 H2O(l) <-> H3O+(aq) + OH-(aq)

and by the usual rules of equilibria:
\[K_w = [H_3O^+][OH^-]\]
where Kw is a constant.

If you add an acid to the solution, it will indeed react with water to produce H3O+, as you observe. But because of the water autoionization equilibrium, any additional H3O+ will force [OH-] down, in order to keep the product [H3O+][OH-] constant and equal to Kw.

Now look back at your original equilibrium. If OH- is added, what does Le Chatelier tell you will happen?

Answer 4

Oops, sorry, I meant if OH- is SUBTRACTED, what does Le Chatelier tell you will happen?

Answer 5

I am not certain at the moment if addition and subtraction behave the same way considering Le Chatelier.

I know if you would add OH-, the balance would shift to the left side (Educts), so is it the opposite if you remove them? Then the balance would shift in favor to the Products.

Answer 6

Yes, correct. Le Chatelier doesn't care whether you add or subtract, the equilibrium always shifts so as to reduce the effect of what you did. If you subtract OH-, the equilibrium will shift so as to increase OH- production, to reduce the effect of your subtraction of OH-.

Answer 7

Okay I understand, thanks a lot!