Question

Question #984b3

Answer 1

Here's how I do it.

Explanation:

First, write the balanced equation for the reaction.

`"Ba(OH)"_2·"8H"_2"O(s)" → "Ba"^"2+""(aq)" + "2OH"^"-""(aq)" + "8H"_2"O(l)"`

`"Moles of Ba(OH)"_2·"8H"_2"O" = 39 color(red)(cancel(color(black)("g"))) × "1 mol"/(315.46 color(red)(cancel(color(black)("g")))) = "0.124 mol"`

`"Moles of Ba"^"2+" = 0.124 color(red)(cancel(color(black)("mol Ba(OH)"_2·"8H"_2"O"))) × "2 mol OH"^"-"/(1 color(red)(cancel(color(black)("mol Ba(OH)"_2·"8H"_2"O")))) = "0.247 mol OH"^"-"`

I presume that the volume of your solution is 1 L. Then

`["OH"^"-"] = "0.247 mol"/"1 L" = "0.247 mol/L"`

`"pOH" = -log["OH"^"-"] = -log(0.247) = 0.61`

`"pH" = "14.00 - pOH" = "14.00 - 0.61" = 13.39`

`["H"_3"O"^"+"] = 10^"-pH"color(white)(l) "mol/L" = 10^"-13.39" color(white)(l)"mol/L" = 4.1 × 10^"-14"color(white)(l) "mol/L"`