Calculate the PH of 0.005 mol dm-3 tetraoxosulphate vi
The formula for tetraoxosulphate VI is SO4^2-. To calculate the pH of a solution of tetraoxosulphate VI with a concentration of 0.005 mol dm^-3, we need to consider the dissociation of water and the dissociation of the tetraoxosulphate VI ion in water.
First, we need to write the equation for the dissociation of water:
H2O ⇌ H+ + OH-
The concentration of H+ ions in pure water at 25°C is 1.0 × 10^-7 mol dm^-3. The concentration of OH- ions in pure water at 25°C is also 1.0 × 10^-7 mol dm^-3. Since water is neutral, the pH of pure water is 7.
Next, we need to write the equation for the dissociation of tetraoxosulphate VI:
SO4^2- + H2O ⇌ HSO4^- + OH-
The equilibrium constant for this reaction is called the acid dissociation constant of the tetraoxosulphate VI ion and is denoted by Ka. Since Ka is very small, we can assume that the concentration of SO4^2- ions is approximately equal to the initial concentration of 0.005 mol dm^-3.
Using the equation for the dissociation of water and the equation for the dissociation of tetraoxosulphate VI, we can write an expression for the concentration of H+ ions in the solution:
Ka = [HSO4^-][OH-]/[SO4^2-]
Since [OH-] = 1.0 × 10^-7 mol dm^-3 and [SO4^2-] = 0.005 mol dm^-3, we can rearrange the equation and solve for [H+]:
[H+] = Ka/[OH-][SO4^2-] = 1.8 × 10^-2 mol dm^-3
Therefore, the pH of the solution of tetraoxosulphate VI with a concentration of 0.005 mol dm^-3 is:
pH = -log[H+] = 1.74