Calculate the pOH of 0.0005moldm^3of tetraoxosulphate vi acid solution
Tetraoxosulphate VI acid is H2SO4, so we need to find the pOH of a 0.0005 moldm^-3 sulfuric acid solution.
First, we need to find the concentration of H+ ions in the solution. Since sulfuric acid is a strong acid, it completely dissociates in water to form H+ ions. Therefore, the concentration of H+ ions is equal to the concentration of the sulfuric acid solution, which is 0.0005 moldm^-3.
Now, we can calculate the pOH using the formula pOH = -log[OH-]. Since the solution is acidic, the concentration of OH- ions is negligible (since H+ and OH- ions react to form water). Therefore, we can assume the concentration of OH- ions is 0.
So, the pOH of a 0.0005 moldm^-3 sulfuric acid solution is calculated as:
pOH = -log(0) = undefined
Therefore, the pOH of a 0.0005 moldm^-3 sulfuric acid solution is undefined.