6.6×10−3 M Ba(OH)2 and 1.00×10−2 M BaCl2 what is [OH-] of solution?
Ba(OH)2 ==> Ba^2+ + 2OH^-
Ksp = (Ba^2+)(OH^-)^2
(Ba^2+) = 6.6E-3 from Ba(OH)2 and 1E-2 from BaCl2.
Solve for (OH^-)
So do you add the 6.6E-3 and 1E-2 together as the value for Ba^2+?
To find the [OH-] in the solution, we first need to determine the concentration of OH- ions based on the dissociation of Ba(OH)2:
Ba(OH)2 → Ba2+ + 2OH-
The concentration of Ba(OH)2 is given as 6.6×10−3 M. Since the ratio of Ba(OH)2 to OH- is 1:2, the concentration of OH- ions is twice that of Ba(OH)2:
[OH-] = 2 × (6.6×10−3) = 1.32×10−2 M
Next, we can determine the concentration of OH- ions in BaCl2 by considering the dissociation of BaCl2:
BaCl2 → Ba2+ + 2Cl-
Since the concentration of BaCl2 is given as 1.00×10−2 M, the concentration of OH- ions is zero because the BaCl2 does not directly produce OH- ions.
Therefore, the overall concentration of OH- ions in the solution is 1.32×10−2 M.
To find the concentration of hydroxide ions ([OH-]) in the solution, we need to consider the dissociation of Ba(OH)2 into its ions.
Ba(OH)2 dissociates into two hydroxide ions (OH-) for every molecule of Ba(OH)2.
Given that the concentration of Ba(OH)2 is 6.6×10^-3 M and the dissociation for Ba(OH)2 is 1:2, the concentration of hydroxide ions can be calculated using the following equation:
[OH-] = 2 × (concentration of Ba(OH)2)
[OH-] = 2 × (6.6×10^-3 M)
= 1.32×10^-2 M
Now let's consider the dissociation of BaCl2 into its ions.
BaCl2 dissociates into one barium ion (Ba2+) and two chloride ions (Cl-) for every molecule of BaCl2.
Since we are looking for the concentration of hydroxide ions ([OH-]) specifically, the dissociation of BaCl2 is not relevant to our calculation.
Therefore, the concentration of hydroxide ions in the solution is 1.32×10^-2 M.