Lakshay Singhal
Last Activity: 11 Years ago
Q. 1
Ans.
pH=2.0, or 2.01 if you want to get more accurate.
I''ll give you the short, simple answer first, then a longer one (but you probably won''t need it).
Sulfuric acid is a diprotic acid. The first proton dissociates from the acid very easily, so you can assume that it will produce 0.005M of H+ ions, or rather H3O+ ions. The second proton does not come off as easily, but to a first approximation, you can assume it does. The second proton generates another 0.005M of H+ ions, so the total [H+]=.010M. Find pH:
pH=-log(0.010) = 2.0
You can do this more rigorously if you need to by not assuming that the second proton completely dissociates. Then you need to calculate the H+ generated by dissociation of the second proton from the 2nd pKa value of sulfuric acid. The chemical equation is
HSO4- (aq) + H2O (l) --> SO4^2- + H3O+ (aq)
The second pKa of sulfuric acid is:
pKa2=1.99, so Ka=10(-1.99)=0.102
Then, we need the equation for Ka2:
Ka2=[SO4^2-][H3O+]/[HSO4-]=0.102
We started with 0.005M HSO4-, since that is what is left after the first proton of sulfuric acid comes off. For every mole of HSO4- that dissociates, you get 1 mole of SO4^2- and 1 mole of H3O+. Set [H3O+]=x and solve the Ka2 equation:
0.102=x^2/(0.005-x)
0.00051 - 0.102x = x^2
x^2 + 0.102x - 0.00051 = 0
Use the quadratic equation to find x:
x={-0.102 +/- sqrt(0.0104+0.00204)}/2
x=0.00478 M = [H3O+] that comes from the 2nd proton.
Total [H3O+] = 0.005 + 0.00478 = 0.00978
Then find pH:
pH = -log [H3O+] = -log (0.00978) = 2.01
So you see our original guess of pH=2.0 is nearly right on.
Q. 2
Ans.
6.96
You add the 10^-8 to the natural 10^-7 for the hydrogen ions.
Then take the negative log of the resulting 1.1x10^-7.
I rounded off to three positions.
Q. 3
Ans.
Please find that out on your own! :D!
