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Topic: Weak Acid and Salt w/ low solubility  (Read 7276 times)

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Makubesu

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Weak Acid and Salt w/ low solubility
« on: April 05, 2005, 06:57:32 PM »
I need help with doing calculations for a lab we did in class.  In the lab, we used a weak acid (white vinegar) and Magnesium Hydroxide.  From the data gathered I already know how much Magnesium Hydroxide was used, and how much vinegar it took to fully dissolve it.  The concentration of acetic acid in vinegar is also known.  The problem is, I'm not sure how to do the calculations to figure out one or the other in order to check the data.  If the acid were a strong acid, I could figure it out pretty easily because there is a known amount of protons to react.  However, it seems that with a weak acid Le Chatelier's principle gets in the way.  Every time the acid reacts with a magnesium, both the acid and the salt get their equilibrium shifted to the right so it changes as the lab goes on.  I'm not sure how to deal with this.

I'm pretty much clueless, and nobody I seem to ask can help me figure it out.  Could someone explain the calculations to me?

Offline Donaldson Tan

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Re:Weak Acid and Salt w/ low solubility
« Reply #1 on: April 05, 2005, 08:32:42 PM »
i believe you are asking how much acetic acid is required to dissolve all Mg(OH)2 present.

you must consider two reactions:
1. CH3COOH + Mg(OH)2 -> (CH3COO)2Mg + 2H2O
2. CH3COOH <-> CH3COO- + H+

Equation (1) gives you a meaure of the required change in concentration of acetate ions in the system. Equation (2) describes the equilibrium system governed by a particular equilibrium constant (Ka).
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Makubesu

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Re:Weak Acid and Salt w/ low solubility
« Reply #2 on: April 05, 2005, 08:45:17 PM »
If I do it like that, though, wouldn't theoretically any salt I used require the same amount of acid?

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Re:Weak Acid and Salt w/ low solubility
« Reply #3 on: April 05, 2005, 08:55:33 PM »
no. if reaction 1 consumes 2 moles of acetic acid, the amount of acetic acid you need to supply is more than 2 moles in virtue of the equlibrium describe by equation (2). the consequence of the equilibrium is that only a certain percentage of the acid present will react with Mg(OH)2.

btw Mg(OH)2 is not a salt.
« Last Edit: April 05, 2005, 08:57:37 PM by geodome »
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Makubesu

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Re:Weak Acid and Salt w/ low solubility
« Reply #4 on: April 05, 2005, 09:05:55 PM »
Thanks for your help so far.  The thing I still don't understand about what you're suggesting though is that it seems like you're saying that the calculations are all reliant on the acid.  But that would mean that I could replace magnesium hydroxide with pretty much any other (are they called ionic solids than?) with two hydroxide groups.  Is that really okay that I do that even if they have differing ksp values?  

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Re:Weak Acid and Salt w/ low solubility
« Reply #5 on: April 05, 2005, 09:12:00 PM »
you can ignore the Ksp calculations because the Ksp equilibrium contributes negligibly to the pH of the system.
"Say you're in a [chemical] plant and there's a snake on the floor. What are you going to do? Call a consultant? Get a meeting together to talk about which color is the snake? Employees should do one thing: walk over there and you step on the friggin� snake." - Jean-Pierre Garnier, CEO of Glaxosmithkline, June 2006

Makubesu

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Re:Weak Acid and Salt w/ low solubility
« Reply #6 on: April 05, 2005, 09:22:11 PM »
Yes and that would be fine if the pH was what I was after.  The problem is, the pH wasn't really involved in this lab.  It appears to have a goal of something like this: you know how much acetic acid it took to completly dissolve the Magnesium Hydroxide.  From this data, determine how much magnesium hydroxide there was.  Or maybe it'd be backwards so it would be more like: you have a known amount of Magnesium Hydroxide and you added acetic acid to it until all of it dissolved.  How much acid did it take?

How would I solve one of those types or problems?  

Edit:
Okay, I believe I came up with a reasonable way to explain what's going on.  I could be completly wrong, but I'm crossing my fingers for a right ;).  Here's a direct quote from my report:

Quote
In this system, Le Chatelier’s Principle provides an intriguing rule.  Let’s begin with the golden rule.  At the point of total dissolving, every proton will have reacted with every hydroxide.  This process does not happen in a set amount of time.  That is because is possible for a proton and hydroxide that are produced to not react with each other until they are once again produced.  This is because the Proton could react with the acetate ion to once again reform acetic acid.  Similarly, the hydroxide could react with magnesium or aluminum.  As the reaction proceeds, the likelihood of the just explained process increases because the ratio of protons/hydroxides to acetate/magnesium/aluminum increasingly favors the acetate/magnesium/aluminum side.
   Here probability arises.  It becomes more and more likely for the proton and the hydroxide not to react, but over an infinite amount of time it still happens.  The ionic solid and the acid continue to keep producing products they run out.  Because in magnesium hydroxide 2 hydroxides are produced per mol of magnesium hydroxide, twice as many moles of acid are needed.

Anyways, if I'm way off, feel free to still say so.  I'm really interested in how to do this type of calculations still, so if anything pops into your memory/mind I'd like to know.
« Last Edit: April 05, 2005, 10:54:48 PM by Makubesu »

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