Question

assuming complete dissociation, what is the pH of a 0.606 M Ba(OH)2 solution? round to 3 significant figures​

Answer 1

Answer:

pH = 14.05.

Explanation:

Hello!

In this case, since barium hydroxide ionizes according to the following equation:

$Ba(OH)\rightarrow Ba^{2+}+2OH^-$

We can compute the concentration of hydroxyl ions based on:

$[OH^-]=0.606\frac{molBa(OH)_2}{L}*\frac{2molOH^-}{1molBa(OH)_2} =1.21M$

Next we compute the pOH:

$pOH=-log([OH^-])=-log(1.12)=-0.0500$

Thus the pH is:

$pH=14-pOH=14+0.05\\\\pH=14.05$

Which means it is very concentrated basic solution.

Best regards!