Question

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

Answer 1

pH = 14.05.

Step-by-step 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(molBa(OH)_2)/(L)*(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!