Calculating pH from Concentration of Strong Acid/Base

[Michelle N - 2C]

On our lecture on January 13th, Lavelle stated on his notes that “If we know the concentration of the strong base/acid, we can calculate pH or pOH.” He then proceeded to the example that 0.0030 M of Ba(OH)2 gives 0.0060 M of OH-. How did that happen? Can someone explain to me this specific process? Thank you!

[nicolely2F]

Since Ba(OH)2 is a strong acid, it is fully dissociated in water. This means that, at equilibrium, zero Ba(OH)2 is left and there is only Ba2+ and OH- in solution, dissociated from the quantity of reactant added to the water.
If you write down the equation for this reaction, you will notice that the coefficients are 1, 1, 2 for Ba(OH)2, Ba2+, OH-. Coupling this with the fact that Ba(OH)2 dissociates 100%, we know that we will have twice as much of OH- as we did of Ba(OH)2. In other words, if we add X mol of Ba(OH)2 to water, the result will be of X mol of Ba2+ and 2X mol of OH-.

[WesleyWu_1C]

When Ba(OH)2 is placed in water, it dissociates into Ba^2+ and 2OH-. So the reaction would be Ba(OH)2 + H2O -> H2O + Ba^2+ + 2OH- For every 1 mole of Ba(OH)2, there are 2 moles of OH-.