Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.700 M Ba(OH)2 solution.
I know that
Ba(OH)2 -> Ba + 2OH
But how do I determine the concentrations? I feel like I am making this more complicated than need be
Yes, you certainly are making it more than it is.
0.700 M Ba(OH)2 means 0.700 mols Ba(OH)2/L soln.
So [Ba(OH)2] = 0.700 M
When it dissociates, according to the equation you wrote, then
(Ba^2+) = 0.700 M
(OH^-) = 2*0.700 M = ? BECAUSE there are twice as many OH^- as Ba^2+. Ba^+ = the same as [Ba(OH)2] BECAUSE one(1) molecule Ba(OH)2 produces one (1) ion of Ba^2+ (and of course two (2) ions of OH^-.
Well, you're definitely in a "Ba-d" situation! Don't worry, I'll help you figure out the concentrations of the individual ions in the Ba(OH)2 solution and we'll "OH-vercome" this together!
Since we have 0.700 M Ba(OH)2, we need to think about how it dissociates. As you correctly pointed out, Ba(OH)2 dissociates into Ba ions and OH ions.
For every 1 molecule of Ba(OH)2, we obtain 1 Ba ion and 2 OH ions. Since the concentration of Ba(OH)2 is 0.700 M, the concentration of Ba ions will also be 0.700 M.
Now, because every 1 molecule of Ba(OH)2 produces 2 OH ions, we need to double the concentration of Ba(OH)2 to get the concentration of OH ions. So, the concentration of OH ions in the solution will be 2(0.700 M), giving us 1.400 M.
I hope this helps clear things up! Just remember, "OH-some" chemistry is all about balancing equations and keeping your ions in check!
To determine the concentration of each individual ion in a 0.700 M Ba(OH)2 solution, you need to understand the dissociation of Ba(OH)2 in water.
Ba(OH)2 dissolves completely in water, yielding one Barium ion (Ba2+) and two hydroxide ions (OH-) for every mole of Ba(OH)2.
Since the molarity (M) of Ba(OH)2 is given as 0.700 M, it means that for every liter (L) of the solution, there is 0.700 moles (mol) of Ba(OH)2.
Step 1: Determine the concentration of Ba2+ ions:
Since one mole of Ba(OH)2 dissociates into one mole of Ba2+ ions, the concentration of Ba2+ ions will also be 0.700 M.
Step 2: Determine the concentration of OH- ions:
Since one mole of Ba(OH)2 dissociates into two moles of OH- ions, the concentration of OH- ions will be twice the concentration of Ba2+ ions. Therefore, the concentration of OH- ions will be 2 * 0.700 M = 1.400 M.
In summary, in a 0.700 M Ba(OH)2 solution, the concentration of Ba2+ ions is 0.700 M, and the concentration of OH- ions is 1.400 M.
To determine the concentration of individual ions in a compound, like Ba(OH)2 in this case, you need to consider the stoichiometry of the compound and its dissociation.
In Ba(OH)2, the ratio of Ba to OH is 1:2, meaning that for every one Ba ion, there are two OH ions. So when Ba(OH)2 dissociates, it will produce one Ba ion and two OH ions.
Given that you have a 0.700 M Ba(OH)2 solution, it means that the concentration of Ba(OH)2 is 0.700 M.
Since Ba(OH)2 dissociates into one Ba ion and two OH ions, the concentration of Ba ions will be the same as the concentration of Ba(OH)2, which is 0.700 M.
The concentration of OH ions will be twice the concentration of Ba ions, as there are two OH ions for every one Ba ion. Therefore, the concentration of OH ions will be 2 * 0.700 M = 1.40 M.
So, in a 0.700 M Ba(OH)2 solution, the concentration of Ba ions will be 0.700 M and the concentration of OH ions will be 1.40 M.