We were given a 0.1 M KOH solution.25cm3 of that solution is taken and 5g of MX precipitate is added to that.X- is an anion of a strong acid.
Kb for MOH=2*(10)^-5
Ksp for MX= 3*(10)^-6 M
1)find the [M+] in the solution
2)Find the [X-] concentration and pH of the solution.
My thoughts on the question:
Reactions taking place in the solution,
MX(s)<==>M+(aq)+X-(aq)-->(1)
KOH--->K+ OH- -->(2)
MX+KOH--->MOH+KX ---(3)
MOH<===>M+ OH- -->(4)
By (ksp/Kb)value we can get the remaining MOH concentration in the solution and then the moles.And we can take the KX moles created is equal to that by the stoichiometry in (3).
And then as we know the initial moles of KOH,we can find the moles reacted.So again by stoichiometry we can say that same value of moles should react from MX too.
But from ksp we get the solubility of MX approximately as 1.7*(10)^-3 moldm-3 and moles of MX as 1.7*25(10)^-6 moles, which seems to be contradiction as the moles reacted is a very bigger value so that can't happen.So do we have to consider all these things.
Or simply from the remaining MOH moles can we apply it in the kb equation and find the remaining M+ concentration? And there can we take [M+]=[X-] or should we apply the value we get as the solubility of MX as the [X-] which is approximately equals to 1.732*(10)^- 3 M?
4 answers
......MX ==> M^+ + X^- Ksp = ....
......OH^- + M^+ ==> MOH 1/Kb =...
-------------------------------
Add to get
MX(s) + OH^- =>MOH + X^-
Keq for this rxn = Ksp/Kb = 0.15
solid...0.1.....0....0 initial
solid...-y.....y.....y change
solid.0.1-y ...y.....yequilibrium
Then (y)(y)/(0.1-y) = 0.15
Solve for y = (M^+) = (X^-) and you should use the quadratic. This gives you (M^+) and (X^-).
(OH^-) = 0.1-y = ? and you convert that to pOH and pH. I believe that will do it.
This question is so far down the list that I almost assumed you had not responded to my first response. If you have further comments please copy and post a new at the top of the page. Thanks. I believe you have one of the best understanding of this concept than any I've seen in the last year or so.