posted by Helena .
Assuming complete dissocaition, what is the pH of 4.93 mg/L BaOH2 solution
There is no BaOH2. You must mean Ba(OH)2
Ba(OH)2 ==> Ba^2+ + 2OH^-
4.93 mg/L = 0.00493 g/L.
mols/L = M = 0.00493/molar mass Ba(OH)2
1 mol Ba(OH)2 gives 2 mols OH^-
Therefore, M OH^- must be twice M Ba(OH)2.
Then pOH = -log(OH^-) and
pH + pOH = pKw = 14. You know pOH and pKw, solve for pH.
Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.800 M Ba(OH)2 solution.
(OH-): ??? M