Sorry - cation and anion!
I mean if we have the following equation:
NaOH(aq) + HCl(aq) <--> Na+(aq) + Cl-(aq) + H2O (l)
we can simply write it as
H+(aq) + OH-(aq) <--> H2O(l)
and calculate deltaH for that equation, as no further reaction happens with the Na+ and Cl- cation/anion?
I assume it's something like [H^+] from the acid reacting with [SO4^2-] from the dissolution of our compound (which reduces the total concentration of [SO4^2-] in solution), which again will increase the amount of [Pb^2+] in solution..?
Ok,
so we have the equilibrium eq. :
10^-2 = [10^-1][SO4^2-] / [HSO4^-].
This results in:
0.1 = [SO4^2-] / [HSO4^-]
I.e. concentration of HSO4^- is significantly higher than SO4^2-. But what exactly does this tell us?
1 mole of water, correct? And then the rest of it must come from the fact that it's a 1L solution?
What about my first question, that relating to the deltaH calculations?
I don't quite understand this. When solving PbSO4(s) we get the equilibrium equation:
PbSO4(s) <--> Pb2+ + SO42-.
I know H2SO4 is a strong acid, that will dissociate completely (or almost completely), Like this:
H2SO4 -> H+ + HSO4-
HSO4- -> H+ + SO42-So as far as I can see, the concentration...
I have a couple of questions related to this task.
The reaction that I proposed was this:
NaOH(aq) + HCl(aq) <--> Na+ + Cl- + H2O (l)
where as the solution manual have this net reaction, as nothing will happen with the Na+ and Cl- ions:
H+ + OH- <--> H2O.
I assume these reactions will...
Ok, so if we have the following reaction:
HA <--> H+ + A-
Then if we decrease the pH - concentration of [H+] will increase, reaction will be shifted towards the left => less products.
If we increase the pH - concentration of [H+] will decrease, reaction will be shifted towards the right =>...