Calculate the pH of 0.005mol/dm³ tetraoxosulphate (vi)
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.
To calculate the pH of a tetraoxosulfate (VI) solution, we need to know its concentration and the dissociation constant (Ka) of the tetraoxosulfate (VI) ion.
Since you mentioned that the concentration of the tetraoxosulfate (VI) solution is 0.005 mol/dm³, we can now proceed with the calculation.
The dissociation equation for tetraoxosulfate (VI) is:
H₂SO₄ ⇌ H⁺ + HSO₄⁻
The dissociation constant (Ka) for tetraoxosulfate (VI) is approximately 1.0 x 10^-2.
First, we need to find the concentration of the H⁺ ions in the solution. Since tetraoxosulfate (VI) dissociates into one H⁺ ion, the initial concentration of the H⁺ ions will be the same as the concentration of the tetraoxosulfate (VI) itself.
[H⁺] = [HSO₄⁻] = 0.005 mol/dm³
Now, we can calculate the pH using the formula:
pH = -log[H⁺]
Let's plug in the values and calculate the pH:
pH = -log(0.005)
Using a calculator, we find that the pH is approximately 2.3.
Therefore, the pH of the 0.005 mol/dm³ tetraoxosulfate (VI) solution is 2.3.