autoionization of water

[jmart]

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So I'm not sure why this concept seems so foreign to me. Could be it has been 7 years since my gen chem course, lol.

Anyways, so in my studying it seems that we can usually calculate pH of acid and bases as long as they are not strong. If they are strong, autionization cannot be ignored. So the methods used to calculate pH will no longer work.

For example, pH of a soln with 1 x 10^-10 mol of HCL to 1L of water.

So using the formula, the answer would be 10 but we know this is not correct, you cannot add acid to a neutral solution and get a more basic solution. So then, it is just said that, 10^-10 is much smaller and won't have an effect on 10^-7. Mathematically, if there were closer, say 10^8, how would you account for this. Is this beyond the scope of the mcat?

So then another question, water aution rxn is endothermic. If temp is increased what happens to pH of solution.

So here we write out the reaction

H2O <--> H+ +OH- (this is not a resonance arrow but my attempt to type an equil arrow)

Ok, it is said that if temp is added, shifts to right. This I understand, a shift the right leads to an increase in H+, which means lower pH. I understand this, but then it says that this shift to the right also leads to a decrease of OH-. This I don't understand, because if both of these are on the product side wouldn't that mean both have an increased concentration? What am I missing here?

 

[loveoforganic]

I'm not strong on top my math tricks and am on an iPod, so will leave the first q for someone else. Second one, you're right. In autoionization, there will definitely be equal quantities of both

 

[cheesier]

So I'm not sure why this concept seems so foreign to me. Could be it has been 7 years since my gen chem course, lol.

Anyways, so in my studying it seems that we can usually calculate pH of acid and bases as long as they are not strong. If they are strong, autionization cannot be ignored. So the methods used to calculate pH will no longer work.

For example, pH of a soln with 1 x 10^-10 mol of HCL to 1L of water.

So using the formula, the answer would be 10 but we know this is not correct, you cannot add acid to a neutral solution and get a more basic solution. So then, it is just said that, 10^-10 is much smaller and won't have an effect on 10^-7. Mathematically, if there were closer, say 10^8, how would you account for this. Is this beyond the scope of the mcat?

So then another question, water aution rxn is endothermic. If temp is increased what happens to pH of solution.

So here we write out the reaction

H2O <--> H+ +OH- (this is not a resonance arrow but my attempt to type an equil arrow)

Ok, it is said that if temp is added, shifts to right. This I understand, a shift the right leads to an increase in H+, which means lower pH. I understand this, but then it says that this shift to the right also leads to a decrease of OH-. This I don't understand, because if both of these are on the product side wouldn't that mean both have an increased concentration? What am I missing here?

To answer the first part of your question, here is how I would think about it. The concentration of OH- and H+ in water is 10^-7. Addition of HCl will protonate the OH-, increasing acidity. Mathematically, you can do it like this:

(10^-7 M OH-) - (10^-10 HCl) = 9.99 * 10^-8 OH- remaining.

10^-10 M OH- was protonated.

-log(9.99 *10^-8) = pOH = 7.00043451

14 - 7.00043451 = 6.999 = pH

I'm not 100% sure that this is correct though, so take it with a grain of salt 😉

 

[ilovemcat]

So I'm not sure why this concept seems so foreign to me. Could be it has been 7 years since my gen chem course, lol.

Anyways, so in my studying it seems that we can usually calculate pH of acid and bases as long as they are not strong. If they are strong, autionization cannot be ignored. So the methods used to calculate pH will no longer work.

For example, pH of a soln with 1 x 10^-10 mol of HCL to 1L of water.

So using the formula, the answer would be 10 but we know this is not correct, you cannot add acid to a neutral solution and get a more basic solution. So then, it is just said that, 10^-10 is much smaller and won't have an effect on 10^-7. Mathematically, if there were closer, say 10^8, how would you account for this. Is this beyond the scope of the mcat?

So then another question, water aution rxn is endothermic. If temp is increased what happens to pH of solution.

So here we write out the reaction

H2O <--> H+ +OH- (this is not a resonance arrow but my attempt to type an equil arrow)

Ok, it is said that if temp is added, shifts to right. This I understand, a shift the right leads to an increase in H+, which means lower pH. I understand this, but then it says that this shift to the right also leads to a decrease of OH-. This I don't understand, because if both of these are on the product side wouldn't that mean both have an increased concentration? What am I missing here?

Anyways, so in my studying it seems that we can usually calculate pH of acid and bases as long as they are not strong. If they are strong, autionization cannot be ignored. So the methods used to calculate pH will no longer work.

Maybe this might clear things up:

Let's say you have a 1M concentration of HCL. HCL, being a strong acid will dissociate completely. Therefore, for every mole of HCL that dissociates, 1 mole H+ is produced, and 1 Cl is produced.

Now let's say you have a beaker of water sitting on a table at 25 degrees celcius. In this beaker, we know that there are precisly, 1x10^-7 H+ Ions and 1x10^-7 OH- ions. If we were to add this strong acid to this beaker of water, the total amount of H+ in the beaker would equal: 1 + 1x10^-7. Because 1x10^-7 is so small, we can simply ignore it and just say our beaker has 1M H+ ions floating around. The 1x10^-7 would barely influence the pH.
So the concentration of our solution is simply: 1M H+

The pH of this solution = -log[H+] or -log[10^0] (because 10^0 = 1M)
Therefore the pH of the solutiohn would equal 0.

Now let's imagine you had a 1x10^-23 M concentration of HCL. Based on what was explained earlier, this is the same thing as saying that, in your solution you have:
1x10^-7 H+ ions PLUS 1x10^-23 H+ ions. Because 1x10^-23 is so small, we can simply ignore that number and say our solution has a concentration of hydrogen ions equal to 1x10^-7 (ie. a neutral pH of 7).

The two above examples were two extreme scenarios, but let's imagine for a moment you have a concentration of HCL nearly equal to the concentration normally found in liquid water at 25 degrees celcius. Let's say your concentration is 10^-8

The total amount of Hydrogen Ions in your solution would equal: 10^-7 H+ PLUS 10^-8 H+
Because the concentrations are CLOSE in value (1 order of magnitude difference), we BOTH concentrations are of equal important and BOTH will equally impact the pH of the solution, but not by much.

Infact, calculating this you'll see that adding 1x10^-8 H+ to a beaker of water (at 25C) will INCREASE the pH of the solution, but only by a little amount:
The pH would change from 7 to 6.95.



Knowing what we know about Equilibrium Constants - A change in temperature would certainly affect the value of Keq. In order to establish how a reaction will respond to this change, you must take into consideration LeChatelier's Principal. For water, were either told or required to know that it's an Endothermic Process. That's essentially the same thing as saying that HEAT is a reactant. (If a reaction was exothermic, HEAT would be a product).

So for water: H20 + Heat --> H+ + OH-

If we increase the temperature what will happen? Well, we'd be adding a reactant (by increasing the temperature). Increasing the temperature is the same thing as adding (more) HEAT to the reactant side (because this is an endothermic process). Now using Le. Chat's principal, how would the reaction respond to this stress? Well, the reaction would want to get rid of some of that heat -- the forward reaction will be favored. If the forward reaction if favored that means there'll be an increase in products and so Keq at this new temperature will increase.

If Kw at 25 degrees celcius is 1x10^-14 -- at a higher temperature, that number must be larger (because more products are produced).
Let's imagine that this value turned out to be 1x10^-13 (1 order of magnitude higher).

That would mean the new pH for a neutral solution of water would equal 6.5 as opposed to a pH 7 (at 25 degrees celcius).



Hopefully this all made sense. 🙂

 

[ilovemcat]

Ok, it is said that if temp is added, shifts to right. This I understand, a shift the right leads to an increase in H+, which means lower pH. I understand this, but then it says that this shift to the right also leads to a decrease of OH-. This I don't understand, because if both of these are on the product side wouldn't that mean both have an increased concentration? What am I missing here?

You would still have equal concentrations of H+ and OH- ions. Increasing the temperature would produce more products - so both would increase, not 1 or the other.

 

[indianjatt]

If i'm not mistaken all of the H+ and OH- from the autoionization of water isn't exactly 1x 10^-7. it's actually slightly less due to the common ion effect, which is why you'd use an ice table, but it's a good approximate to use 1 x 10^-7 M OH-/H+ for the autoionization of water on examns when your concentration of Acid/Base is close to the magnitude of 10^-8 or so (where the autoionization of water is significant).