Calculate the pH of a 4.00 x 10^-2 mol dm^-3 solution of Ba(OH)2

The easiest way to answer this question is by using the equation pH + pOH = 14. 

Firstly, we need to work out the number of moles of OH^- in the solution. As the formula for Ba(OH)₂  shows that it has 2 moles of OH^- for every 1 mole of Ba⁺ , the number of moles of OH^- ions will be equal to twice the concentration of the solution. 

Therefore, there are 8.00 x 10^-12 moles of OH^- ions in the solution.

As pOH= -log (OH^-), pOH = -log (8.00 x 10^-12) 

Therefore pOH= 1.097

By rearranging the formula pH + pOH = 14, we get 14-pOH= pH

Therefore, 14-1.097= 12.9

The pH of the solution is 12.9