We need to know the dissociation of tetraoxosulphate (vi) ion, also known as sulphuric acid (H2SO4):
H2SO4 ⇌ H+ + HSO4-
Since HSO4- is still an acid, we need to dissociate it again:
HSO4- ⇌ H+ + SO4²-
Now we can write the equilibrium expression for the dissociation of H2SO4:
Kw = [H+][HSO4-] / [H2SO4]
where Kw is the ion product constant of water (10^-14)
Since we know that [H2SO4] is 0.005 mol/dm³, we can assume that [H+] = [HSO4-] (because HSO4- does not dissociate appreciably at this concentration). Therefore:
Kw = [H+]² / 0.005
Solving for [H+], we get:
[H+] = √(Kw x 0.005)
[H+] = √(10^-14 x 0.005)
[H+] = 7.07 x 10^-8 mol/dm³
Taking the negative logarithm of [H+], we get the pH:
pH = -log[H+]
pH = -log(7.07 x 10^-8)
pH = 7.15 (rounded to two decimal places)
Therefore, the pH of 0.005mol/dm-³ tetraoxosulphate (vi) is 7.15.
Calculate the pH of 0.005mol/dm-³ tetraoxosulphate (vi)
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