Calculate the pH of 0.005moldm-3 tetraoxosulphate(VI)

Answer 1

I assume you mean H2SO4. The correct IUPAC name is sulfuric acid.
Is this a beginning or an advanced class?
Beginning:
...............H2SO4 --> 2H^+ + SO4^2-
(H^+) = 2*0.005 = 0.01
pH = -log 0.01 = 2
Advanced:
H2SO4 is a diprotic acid. The first H^+ is a strong acid but the second H^+ is weak.
..............H2SO4 ==> H^+ + HSO4^- complete ionization so H^+ = 0.005 M
..............0.005.......0.005.....0.005
Then HSO4^- ==> H^+ + SO4^2- for which k2 = 0.012
I.........0.005...........0.005.............0
C..........-x...............x..............x
E........0.005-x.....0.005+x.......x
You need to use the k2 in you text/notes. My text says 0.012
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Substitute (0.005+x)(x)/(0.005-x). Solve the quadratic and final (H^+) = 0.005 + x. Then convert to pH. I obtained 0.003 for the second ionization so total H^+ = 0.005+0.003 = 0.008 for pH = 2.1
Compare the two sets; i.e., beginning H^+ = 0.01 and pH 2 with advanced H^+ = 0.008 and pH 2.1.

Answer 2

I don't understand

Answer 3

I am lost

Answer 4

Calculate the ph of h2so4 with 0.005mol/dm3

Answer 5

Calculate the PH of 0.01M tetraoxosulphate vi acid solution

Answer 6

The correct formula for tetraoxosulphate(VI) acid is H2SO4.

To calculate the pH of a 0.01M H2SO4 solution:

H2SO4 dissociates in water as follows:

H2SO4 → 2H+ + SO4^2-

The concentration of H+ ions in the solution is 2 x 0.01 = 0.02 M.

pH = -log[H+] = -log(0.02) = 1.7

Therefore, the pH of a 0.01M H2SO4 solution is 1.7.