what is the molarity of the individual ions in a 5.0E-5M solution of barium hydroxide?
To find the molarity of the individual ions in a solution of barium hydroxide (Ba(OH)2), we first need to understand the dissociation of the compound in water.
Barium hydroxide dissociates into one barium ion (Ba2+) and two hydroxide ions (OH-) according to the following balanced equation:
Ba(OH)2 → Ba2+ + 2OH-
In a 5.0E-5M (5.0 x 10^-5 M) solution of barium hydroxide, the overall concentration given refers to the concentration of the barium hydroxide as a whole.
To determine the molarity of the individual ions, we need to take into account the stoichiometry of the dissociation. From the balanced equation we can see that for every 1 mole of barium hydroxide, we generate 1 mole of barium ions and 2 moles of hydroxide ions.
Therefore, the molarity of the barium ions (Ba2+) will also be 5.0E-5M, and the molarity of the hydroxide ions (OH-) will be twice that, which is 2 * 5.0E-5M = 1.0E-4M.
To summarize:
- The molarity of barium ions (Ba2+) in a 5.0E-5M solution of barium hydroxide is also 5.0E-5M.
- The molarity of hydroxide ions (OH-) in the same solution is 1.0E-4M.