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Calculate the OH- concentration of 0.01m tetraoxosulphate(vi) acid

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To calculate the OH- concentration of a 0.01M tetraoxosulphate(vi) acid (H2SO4), we first need to determine the pOH of the solution.

Since tetraoxosulphate(vi) acid is a strong acid, it completely dissociates in water:
H2SO4 -> 2H+ + SO4^2-

This means that the concentration of H+ ions is equal to the initial concentration of the acid, which is 0.01M. The pOH is calculated as:
pOH = -log[H+]
pOH = -log(0.01)
pOH = 2

Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
pH = 14 - 2
pH = 12

Now that we have the pH of the solution, we can calculate the concentration of OH- ions using the equation:
[OH-] = 10^(-pOH)
[OH-] = 10^(-2)
[OH-] = 0.01M

Therefore, the OH- concentration of a 0.01M tetraoxosulphate(vi) acid is 0.01M.