what is the [h3o+] in a 9.93 x 10^-6 M Ba(OH)2 solution?

(this is all that's given)

I understand the basics of the question. however does [OH]= 9.93x10^-6 squared?

No, but you're on the right track. Ba(OH)2 = 9.93E-6 from the problem. Since there are two OH ions in 1 Ba(OH)2, then OH = 2*9.93E-6. Then (H3O^+)(OH^-) = 1 x 10^-14

To determine the [H3O+] in a solution of Ba(OH)2, we need to consider the ionization of water and the dissociation of Ba(OH)2.

Ba(OH)2 is a strong base, meaning it dissociates completely in water. The balanced chemical equation for the dissociation of Ba(OH)2 is:

Ba(OH)2 -> Ba2+ + 2OH-

In this case, the concentration of Ba(OH)2 is given as 9.93 x 10^-6 M.

Since Ba(OH)2 dissociates into two OH- ions, the concentration of OH- ions in the solution will be twice the concentration of Ba(OH)2. Thus, the concentration of OH- ions is calculated as:

[OH-] = 2 × (9.93 x 10^-6 M) = 1.986 x 10^-5 M

Water also has a tendency to ionize into H+ and OH- ions. However, in pure water, the concentration of H+ ions from self-ionization is usually very low, around 1.0 x 10^-7 M.

H2O -> H+ + OH-

Since Ba(OH)2 is a strong base, it will react with water to consume OH- ions and produce more H+ ions. For every two OH- ions consumed, two H+ ions are produced. This means the increase in the concentration of H+ ions will also be twice the decrease in OH- ions concentration.

Therefore, the [H3O+] concentration is:

[H3O+] = 2 × (1.0 x 10^-7 M - 1.986 x 10^-5 M)
= 2 × (-1.976 x 10^-5 M)
= -3.95 x 10^-5 M

Thus, the [H3O+] in a 9.93 x 10^-6 M Ba(OH)2 solution is approximately -3.95 x 10^-5 M.

Bob Pursley showed you how to work the problem. His question was designed to help you work the problem. You don't need any other information. Plug in OH and Kw and calculate H3O^+