6A.23

[Reina Robles 2B]

"6A.23 Calculate the molar concentration of Ba(OH)2(aq) and the molar concentrations of Ba21, OH2, and H3O1 in an aqueous solution that contains 0.43 g of Ba(OH)2 in 0.100 L of solution."

Are we supposed to use the ice box method to solve this? K wasn't given.

Thanks!

[nicolely2F]

You don't need K nor the ICE table because Ba(OH)2 is a strong base and thus will be 100% dissociated in water. This means you can assume the equilibrium concentration of the reaction will be:
[Ba2+]final = [Ba(OH)2]initial = 0.025M
and
[OH-]final = 2*[Ba2+]final = 0.05M
Remember that these conclusions are drawn from the coefficients in the balanced dissociation reaction of Ba(OH2), and that 0.43g of Ba(OH)2 = 0.0025M (divide that by 0.100L to obtain the concentration of 0.025M).

Then, you can use the Kw equation and plug in [OH-] to find [H3O+].

[Justin Vayakone 1C]

Ba(OH)2 is a strong acid, so you can assume all of the reactant turns into products. No need for an ICE table or K value. Just be careful and know that there will be double [OH] compared to [Ba2+].

[Reina Robles 2B]

You don't need K nor the ICE table because Ba(OH)2 is a strong base and thus will be 100% dissociated in water. This means you can assume the equilibrium concentration of the reaction will be:
[Ba2+]final = [Ba(OH)2]initial = 0.025M
and
[OH-]final = 2*[Ba2+]final = 0.05M
Remember that these conclusions are drawn from the coefficients in the balanced dissociation reaction of Ba(OH2), and that 0.43g of Ba(OH)2 = 0.0025M (divide that by 0.100L to obtain the concentration of 0.025M).

Then, you can use the Kw equation and plug in [OH-] to find [H3O+].

Ba(OH)2 is a strong acid, so you can assume all of the reactant turns into products. No need for an ICE table or K value. Just be careful and know that there will be double [OH] compared to [Ba2+].

Thank you both so much!