Tetraoxosulphate (VI) is commonly known as sulphuric acid (H2SO4). It is a strong acid with a dissociation constant (Ka) of 1.2 x 10^-2.
To calculate the pH of 0.005mol/dm³ sulphuric acid:
1. Use the dissociation equation for H2SO4:
H2SO4 + H2O ⇌ HSO4- + H3O+
2. Use an ICE (initial, change, equilibrium) table to determine the equilibrium concentration of H3O+:
H2SO4 + H2O ⇌ HSO4- + H3O+
I 0.005mol/dm³ 0 0
C -x +x +x
E 0.005-x x x
The equilibrium concentration of H3O+ is x.
3. Use the equilibrium concentration of H3O+ to calculate the pH:
pH = -log[H3O+]
pH = -log(x)
From the ICE table, we know that x = 1.41 x 10^-2 mol/dm³.
pH = -log(1.41 x 10^-2)
pH = 1.85
Therefore, the pH of 0.005mol/dm³ sulphuric acid is 1.85.
Calculate the pH of 0.005mol/dm³ tetraoxosulphate (vi)
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