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.

Question

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?

Answer 1

I had trouble following your thoughts after the first few lines but the first part is OK. You started out right. You need Ksp/Kb = 0.15
......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.

Answer 2

What about the concentration of KX ,formed from the reaction of KOH and MX? Doesn't KX dissociate and do we have to neglect the hydrolysis as X- is a strong acid anion?

Answer 3

y gives the concn of X^- and therefore of KX. And X^- is not hydrolyzed for the reason you state.
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.

Answer 4

I see I wrote y = (M^+) and that isn't right. y = (MOH) = (X^-). You know Kb for MOH and you know OH so you can calculate (M^+).