Question

What is the pH of a 0.050 M Ba(OH)2 aqueous solution?

Answer 1

First, let's see that we can extract the concentration of OH- because Ba(OH)2 is a base. Let's see the dissociation of this base:

$Ba(OH)_2\to Ba^{2+}+2OH^-,$

You can realize that we have 2 moles of OH-. The next step is to multiply this number of moles by the concentration (0.050 M):

$\lbrack OH^-\rbrack=2\cdot0.050=0.1.$

Remember that the formula of pOH is -log ( [OH-] ):

$\text{pOH}=-\log (0.1)=1.$

And with this result, we can find pH, using the formula:

$pH+\text{pOH}=14$

And we're going to obtain:

$\begin{gathered} pH=14-\text{pOH}, \\ pH=14-1, \\ pH=13. \end{gathered}$

The pH of the solution would be 13, so the answer is (2).