Calculate the pH of 0.005mol/dm-³ tetraoxosulphate (vi)
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.
To calculate the pH of a solution of tetraoxosulphate (VI), also known as sulfuric acid (H2SO4), we need to consider the dissociation of the acid into its ions.
The equation for the dissociation of sulfuric acid is:
H2SO4 ↔ 2H+ + SO4^2-
Sulfuric acid is a strong acid, which means it completely ionizes in water. This means that for every mole of sulfuric acid that dissociates, it releases 2 moles of H+ ions.
In this case, we have 0.005 mol/dm^3 of sulfuric acid. Since it completely ionizes, the concentration of H+ ions is also 0.005 mol/dm^3.
To calculate the pH, we use the formula:
pH = -log[H+]
Therefore, pH = -log(0.005) = 2.3
So, the pH of a 0.005 mol/dm^3 solution of tetraoxosulphate (VI) is 2.3.