[sameeralord]

Hello everyone,

My head is currently overflowed with dynamic equilibrium . Anyway now that I'm over that I got a question on acid base equilibrium,

a)Calculate PH of 0.10 M Mg(OH)₂ solution at 25 degrees Celsius

Ok I know how to do this but my question is in the first step. Where we work out the concentration of OH using mole ratios.

Mg(OH)₂ -------> Mg²⁺ + 2OH^-

Our teacher has used mole ratios to work out the concentration.  2x0.10. I thought we only use mole ratios to work out moles. Can we use mole ratios to work out concentraion. I also want to know if the solution is basic. The answer is 13.3. But I'm confused because in the self ionisation of water the solution is neutral even though PH is high or less. To tell you the truth I don't understand why this is a self ionisation of water question.

b) What would happen to the PH if the temperature rose to 35 degrees Celsius?

Our teacher has used 2H₂0 <-----> H₃0⁺ + OH^-  Change in h= +57 Kj mol^-  and Le chatelier to solve this.

I can understand how the Le Chatelier works in this equation but my question is why does the teacher relate this water formula to almost all the acid base questions. Isn't the formula for this question what I wrote in first question not the water one.

Even for a question like this our teacher has used the formula
2H₂0 <-----> H₃0⁺ + OH^-
0.2 L of a solution of PH is added to 80 ml of 0.010 M NaOH. Calculate the PH of the resulting solution.

Why is that we can use this water formula for every acid base question. Any help would be appreciated. Thanks

[sameeralord]

OK I think I quite worked the water problem my self. Mg(0H)₂
is dissolved in water. Hence this increases the concentration of OH^- in water. So does this mean the system moves left to partially oppose the change. So there would be less H₃0 ⁺ and slightly more OH^- ions than it began with(due to partial change). Leading to a basic solution. Am I right?

In acid base neutralisations I have a problem. If water is produced by this reaction wouldn't this make the equilibrium position to move to the right. Does this happen in acid base equilibrium?

[cliverlong]

Hi,

   Without having read your question in detail, any question pH related I would go straight to Borek's ChemBuddy web site

http://www.chembuddy.com

then select pH lectures

Also do a search on "pH" and equilibrium" in this forum and I'm sure you will find hundreds of hits.

I suggest you work your way through a few of those and see if anything is relevant to your question.

I think I created a post somewhere on this forum (maybe I was dreaming) that tried to simply explain the difference between concentration and weak/strong acids and alkalis - and the effect that had on pH. I didn't write anything new or novel - just drew together existing info from the net in one place and in a style that made sense to me. Not sure if that's relevant to yoru question. I'm just a bit rushed at the moment otherwise I would read your question a bit closer.

Good luck, let us know how you get on.

Clive

[cliverlong]

. but my question is why does the teacher relate this water formula to almost all the acid base questions. Isn't the formula for this question what I wrote in first question not the water one.

Even for a question like this our teacher has used the formula
2H₂0 <-----> H₃0⁺ + OH^-
0.2 L of a solution of PH is added to 80 ml of 0.010 M NaOH. Calculate the PH of the resulting solution.

Why is that we can use this water formula for every acid base question. Any help would be appreciated. Thanks

I will have a crack at that part (but I'm not sure if I am right).

Using Bronsted-Lowry definition of acids and bases

An acid is a proton donor
A base is a proton acceptor.

Now, the acid has to have something to give the proton to

In the case of hydrogen chloride, we can use water. As disassociation is high, acid is strong. Water acts as a base.

Similarly, if sodium hydroxide is dissolved in water, there is a significant (vast?) excess of hydroxide ions. Because water is only slightly disassociated, the hydroxide binds with free hydrogen ions to form water. Hence concentration of H⁺ drops and pH goes up.

So water has acted as either an acid or a base depending on the other chemical.

So where does the 2H₂0 <-----> H₃0⁺ + OH^- come into this?

Maybe as follows.

The K_w value at 25^oC is very nearly 10^-14

see here >> http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product for much more on this

This K_w is our "reference point" leading to a pH for water (depending on temperature) of around 7

Now, when you want to calculate the pH of a substance dissolved in water you need to:

calculate concentration of solution
be given the disassociation constant for the substance K_a or K_b

Now ... the only use I have found for K_w in such calculations is trying to determine the pH for a strong base. Consider sodium hydroxide dissolved in water. Assume all the OH^- comes from the sodium hydroxide. So we can calculate [OH^- ]
We convert this to [H⁺] by observing [H⁺] = K_w / [OH^- ] and then we get pH. They may be other uses for K_w I haven't thought of.

Since many ionic substances dissolve in water, and as shown above water can act as either a proton donor (acid) or a proton acceptor (base), water is a very useful medium in which to create acids and alkalis of required pH.

Other solvents, if they can act as proton donors and / or acceptors can substitute for water.

Water is not needed if the acid and base can react directly, such as gaseous hydrogen chloride and ammonia.



Clive