3. Potassium hypomanganate, K3MnO4, is a bright blue salt and a rare example of

Question

3. Potassium hypomanganate, K3MnO4, is a bright blue salt and a rare example of a manganese(V) compound. MnO43 is not stable and readily disproportionates according to the reaction 2 MnO43-(m) + 2 H2O(i) MnO4-(ag) + MnO2(s) + 4 OH-(aa) sa= 4.61 x 1011 a) What pH is necessary to maintain a hypomanganate concentration of at least 5.75 x 104 M if the solution is 0.405 M KMnO4? b) Given that the maximum solubility of potassium hydroxide is 21.5 M, could this concentration of hypomanganate be maintained?

Explanation / Answer

a) According to law of mass action, (for reverse rxn)

k=1/Keq=[MnO43-]^2/[MnO4-][OH-]^4 [concentration of pure solid and pure liquid does not change appreciably at equilibrium so ,not included in equilibrium constant expression]

ICE table

k=1/keq=1/(4.61*10^11)=2.169*10^-12

k=2.169*10^-12=([OH-]o-4x)^4 (0.405M)/(5.75*10^-4M)^2

[OH-]o-4x=3.648*10^-5M=equilibrium concentration of OH-

[H+]=kw/[OH-]=10^-14/(3.648*10^-5)=2.741*10^-10

pH=-log[H+]=-log (2.741*10^-10)=9.6

pH=9.6

b) if [OH-]eq=21.5M

k=2.169*10^-12=([OH-]o-4x)^4 (0.405M)/([MnO43-]o-2x)^2

where

[MnO43-]o-2x)=eqm concentration of hypomanganate

([OH-]o-4x)=eqm con of OH-=21.5M

So,2.169*10^-12=(21.5M)^4 (0.405M)/[hypomanganate]^2

or,[hypomanganate]=5.006*10^-9 M <<< (5.75*10^-4M),so this concentration could not be maintained

[MnO4-] [MnO43-] [OH-] initial 0.405M [MnO43-]o [OH-]o change -x +2x -4x equilibrium 0.405-x=0.405M(x<<0.405M) [MnO43-]o+2x =5.75*10^-4M [OH-]o-4x

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