Inorganic chemistry pH question?

Question

What is the pH of a 5.0 M H3PO4solution?
ka1=7.5 x 10-3
ka2=6.2 x 10-8
ka3=4.8 x 10-13

Answer 1

You have an extremely high concentration (5M) which would require that you use a method to calculate activity coefficients in order to calculate accurately the pH. Since Ka1=7.5*10^-3 you would have approximately [H+]=[H2PO4(-)]= 0.1936 M which means that the Debye-Huckel method is not good enough and you probably have to use the Pitzer model. This is beyond my abilities so I won't get into that.

Let's try to find an approximate solution.
The least accurate approximation is to assume that we have only the first dissociation event (justified by the fact that Ka2 is 5 orders of magnitude smaller) and that the amount of H3PO4 at equilibrium is practically the same as the initial (that is the amount x that dissociates is <<5 and 5-x=5) then we get the answer above which is x=0.1936 and
pH=-logx =-log(0.1936)=0.71

Well, 0.19 is not <<5 so our assumption is not valid. We need to solve the quadratic that comes from Ka1
Ka1=x^2/(5-x)=>
x^2+0.0075x-0.0375=0
x=0.18994
and pH=-logx =-log(0.18994) = 0.72

If you want to take the first 2 dissociation events in consideration then
(look at http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base for deriving the formula)

[H+]^2-(C*(Ka1[H+]^2 +2*Ka1*Ka2[H+]) / ([H+]^2 +Ka1[H+]+Ka1Ka2) -Kw=0

solve that and [H+]=0.189936. You see that it is practically the same as when you ignored the second dissociation, so you can easily get away with ignoring the second and third.

You need to do a correction for the ionic strength. However I don't know exactly your background and the requirements of the problem you posted. The best model would be the Pitzer, but it is not something I know (and most people don't know that). If your teacher has taught you other models (like the Debye-Huckel or extended Debye-Huckel) apply the one you have learned (though in reality they are not accurate at such high concentrations, assuming that the only species in solution contributing to ionic strength are H+ and H2PO4(-) with [H+]=[H2PO4(-)] =0.18994 M.

Answer 2

In situation a million, there is an errors; Kb= 2x10^-5. CN- acts as a base" CN- + H2O -> HCN + OH- So if we initiate with a nil.01 M answer, a number of it, say x moles/L will react to type x moles of HCN and x moles of OH- ion. Then...................... x^2 / 0.01-x = 2x10^-5 x=4.5 x 10^-4. this implies you have dropped (x) whilst in comparison with 0.01 without sacrificing too lots accuracy. 2.2 x 10^-10 is the H+ ; from [H+][OH-] = 10x10^-15. partly b. your chem rxn seems in errors; I desire MeNH2 + H2O -> MeNH3+ + OH- , which shows the hydroxyl formation . this could be a buffer answer situation in cover, and the conc of the susceptible base and its acid salt are severe adequate that the quantity reacted to realize equilibrium is small in assessment. Then........ 0.15 x [OH-]/0.20 = 4.4x10^-4 and [OH-] = 6x10^-4 appx. From the H2O equilibrium above, [H+] = a million.7x10^-11 appx, and the pH would be approximately 10.seventy 5