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?
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.
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?
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.
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^+).
To answer the questions:
1) To find the concentration of M+ in the solution, we can start by determining the concentration of OH- ions in the solution.
We are given that the initial concentration of KOH is 0.1 M, and we need to consider the reaction (2) between KOH and water:
KOH --> K+ + OH-
Since KOH is a strong base, it fully dissociates in water, meaning that the initial concentration of OH- ions is also 0.1 M.
Next, we need to consider the reaction (4) between MOH and water:
MOH <==> M+ + OH-
The equilibrium constant for this reaction is Kb. We are given that Kb = 2*(10)^-5.
We can use the equation for Kb to determine the concentration of M+ ions:
Kb = [M+][OH-]/[MOH]
Since [OH-] is the same as the initial concentration of KOH, which is 0.1 M, and [MOH] is the concentration of MOH that remains after the reaction (3), we can rearrange the equation to solve for [M+]:
[M+] = Kb * [MOH] / [OH-]
Therefore, we need to determine the concentration of MOH remaining in the solution after the reaction (3).
2) To find the concentration of X- and the pH of the solution, we need to consider the reaction (3) between MX and KOH:
MX + KOH --> MOH + KX
We are given that the Ksp for MX is 3*(10)^-6 M.
The Ksp expression for MX is:
Ksp = [M+][X-]
We can use this equation to determine the concentration of X- ions.
Since we know the initial concentration of MX is 5g and we can calculate the number of moles of MX based on its molar mass, we can then calculate the concentration of X- using the equation:
[X-] = [X-] remaining in solution = sqrt(Ksp)
Then, to find the pH of the solution, we need to calculate the pOH first. The pOH is given by:
pOH = -log10[OH-]
Since we already know the concentration of OH- from the initial concentration of KOH, we can calculate the pOH.
Finally, we can use the relation between pH and pOH:
pH + pOH = 14
Therefore, to find the pH of the solution, we can subtract the pOH from 14.