Determine the pH of a .50 M solution of H2S04 and also determine the concentration of the sulfate ion (SO42-)
I don't know where you are in chemistry but this problem is not as easy as it looks. The problem is that H2SO4 is a strong acid (100% ionized) for the first H^+ BUT the second one is not completely ionized (k2 = 1.2E-2). As a result, the first H that comes off reduces the ionization of the second one even further so the pH is essentially determined by the first ionization. ]
H2SO4 ==> H^+ + HSO4^-
HSO4^- ==> H^+ + SO4^2-
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
So you set up an ICE chart for k1 and k2.
..............H2SO4 ==> H^+ + HSO4^-
initial.......0.5M0......0......0
equilibrium.....0.......0.50....0.50*see below.
..............HSO4^- ==> H^+ + SO4^2-
initial......0.50.......0.......0
change.........-x.......x.......x
equil........0.50-x......x.......x
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Now substitute.
(H^+) 0.50 from first equn + x from second.
(SO4^2-) = x from second equn
(HSO4^-) = 0.50-x frm second equn.
0.012 = (0.50+x)(x)/(0.5-x) and solve for x. You will need to solve the quadratic OR you can do it by successive approximations.
The (H^+) then is 0.5+x and x is the sulfate.
In the above I used k1 after saying H2SO4 had no k1. It doesn't. I should have said the first ionization which I referred to k1.
To determine the pH of a solution of H2SO4, we need to consider its dissociation in water. H2SO4 is a strong acid that ionizes completely in water.
The dissociation of H2SO4 can be represented by the equation:
H2SO4 → 2H+ + SO42-
Since H2SO4 dissociates into hydronium ions (H+) and sulfate ions (SO42-), the concentration of H+ ions would be twice the concentration of H2SO4, which is 0.50 M in this case.
Therefore, [H+] = 2 * 0.50 M = 1.00 M
The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions. Thus, we can use the formula:
pH = -log[H+]
pH = -log(1.00)
pH ≈ -0.00 (rounded to two decimal places)
Therefore, the pH of the 0.50 M solution of H2SO4 is approximately 0.00.
To determine the concentration of sulfate ions (SO42-) in the solution, we can directly use the stoichiometry of the dissociation equation. For every 1 mole of H2SO4, we get 1 mole of SO42- ions.
Therefore, the concentration of SO42- ions in the solution is the same as the concentration of H2SO4, which is 0.50 M.