00:01
To answer this question, we can do the same type of calculations as would be done in the previous problem.
00:08
This would require us to know this particular chemical reaction, where as in the book, it describes the metal sulfide reacting with the hydronium ion present in the acidic solution, producing the metal and the hydrogen sulfide.
00:29
They call the equilibrium then a k -s -p -a, which is a ratio.
00:34
Of the metal ion concentration, multiplied by the hydrogen sulfide concentration, divided by the hydronium concentration squared.
00:41
Turns out that this same expression is the qc expression if we do not know if these concentrations are equilibrium concentrations.
00:50
What we then have to do is after calculating the qc value, we compare it to the equilibrium k -spa value, which in this case, the metal being iron 2 plus, was given to us as...
01:04
1 .0, no, for the iron 2 plus sulfide as 6 times 10 to the 2.
01:12
So to carry out the k, sorry, the qc calculation for the iron 2 plus that we have, we know that we have a 0 .1 molar concentration of the metal and then a 0 .1 molar concentration of h2s.
01:28
The hydronium concentration is given to us in this very acidic solution at 0 .4 molar.
01:38
Then carrying out the calculation, we get qc is equal to 6 .25 times 10 to negative 2.
01:47
Because for iron sulfide, the kspa value is 6 times 10 to the 2, then we know that the qc value is less than the kspa value.
02:06
Therefore, when this is the scenario, the reaction ends up shifting this way, and we do not get the metal to precipitate, because we need it to go to the left in order to get the metal to precipitate...