pH Calculation

Hiya,
I need help, as I do not understand much of my chemistry calculationas at the moment. I would appreciate some help as I really am not sure how to work out some of these problems.

1) Is the dissociation of water endothermic or exothermic? Justify answer by referance to the given values of Kw.
2) pH of water is 7.47 at 273K, calculate Kw at this temperature.
3) Calculate the hydrogen ion concentration and pH of a solution of propanoic acid containing 0.20 mol. (Ka=1.35x10-5)
Lastly...
4) Calculate the pH when 10cm of 0.1 mol HCl are added to 990cm of water.

I would appreciate some help as I have really tried my best to try and figure out how to work these out, but I just dont know what to do or where to start... If someone could give me a perfect method of answering these type of questions, I will know what to do in future.
Thanks.

Reply 1

selkie222

Sorry haven't done much chem in the hols... so I might be a bit slow. Have you done any of these questions or gone over any of the points before in lesson? These are not very tricky questions and you should be worried if your teacher doesn't tell you anything about them!

1) Is the dissociation of water endothermic or exothermic? Justify answer by referance to the given values of Kw.

I presume you are given a list of Kw values against temperature?

From that list, you will probably spot that as temperature increases, Kw increases.

Now remember the definition of Kw:
Kw = [H+][OH-]
ie Kw is the concentration of H+ multiplied by the concentration of OH-

The increasing Kw values suggest that as temperature increases, both [H+] and [OH-] also increase.

Now the dissociation reaction is actually an equilibrium, ie
H2O <--> H+ + OH-

Remember that an equilibrium system will try to oppose a change/constraint. In the case of increasing temperature, it should shift to the endothermic direction to oppose the change.

As we've seen above, if temperature increases, [H+] and [OH-] also increases. This suggests that the forward reaction is endothermic.

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Another way of thinking about it without using Kw values

Think dissociation means
H2O --> H+ + OH-
in this process you are breaking the bond H---OH, and breaking bonds require energy, so it is endothermic.

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2) pH of water is 7.47 at 273K, calculate Kw at this temperature.

As we've said
Kw = [H+][OH-]

The question doesn't mean that the water is alkaline, it is still "neutral" water with [H+] = [OH-] The reason that pH is not 7.00 even though the water is neutral I'll explain further on.

So for this calcuation,
pH = -log[H+]
=> [H+] = 10^(-pH) = 10^(-7.47)

and since [H+] = [OH-],
[H+] = 10^(-7.47)
[OH-] = 10^(-7.47)

and
Kw = [H+][OH-]
= 10^(-7.47) * 10^(-7.47)
= 10^(-14.94)
= 1.15 * 10^(-15) mol2dm-6

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pH of water is only 7.00 at 298K. As temperature goes up, [H+] and [OH-] also goes up as we have discussed above. Thus, [H+] will become greater and pH will become lower.

High [H+] (usually) corresponds to acidity and low pH, if it helps.

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3) Calculate the hydrogen ion concentration and pH of a solution of propanoic acid containing 0.20 mol. (Ka=1.35x10-5) I think that's containing 0.20moldm-3...?

I'll use HA to represent propanoic acid, as in
HA <--> H+ + A-
and is analogus to
CH3CH2COOH <--> H+ + CH3CH2COO-

Using
HA <--> H+ + A-
We could define Ka just like any other equilibrium constant---
[H+][A-] = Ka
[HA]

Now since we are dealing with a "weak acid situation" (as my teacher used to call it), we could make certain approximations to help ourselves.

A weak acid only dissociates slightly in solution. So, we could presume that [HA] in the equilibrium expression is the same as the total concentration of HA. In this case [HA]total = 0.20, after dissociation it might be 0.1999995 but we'll say that it won't make too much of a difference if we use 0.20 for our calculation.
From the dissociation eqation, HA <--> H+ + A- we could see that [H+] = [A-]

Now we have the approximate expression:
[H+]^2 approximately = Ka
[HA]total

Now we could plug in the numbers...
Ka = 1.35x10-5
[HA]total = 0.20

[H+] = sqrt(Ka[HA])
= sqrt(2.7x10^-6)
= 1.64^10-3 moldm-3

And from this pH is easy:
pH = -log[H+]
= -log(1.64^10-3)
= 2.78

ok?

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4) Calculate the pH when 10cm of 0.1 mol HCl are added to 990cm of water.

moles of HCl = 0.1*10/1000 = 0.001moles
volume of liquid = 10 + 990 = 1000cm3 = 1.00dm3

Assuming HCl dissociates completely, moles of HCl = moles of H+

[H+] = 0.001/1 = 0.001 moldm-3

pH = -log(0.001)
= 3.00

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Hope this helps. If it's too late, I suppose I could console myself that I've revised this bit of chemistry before I go to uni! ^_^

Cheers,
selkie

Reply 2