c=Calculate the OH- conc in a solution containing 6.2x10(-3)M Ba(OH)2 and 1.00x10(-2)M BaCl2?
I thought of using the BaCl2 conc to find the conc of Ba initially, then use that in the ice table for Ba(OH)2, and solve for OH. But i dnot think that's right.
Help? :/
I wonder if this problem expects us to use a Ksp for Ba(OH)2 or not. If not I see no reason to have included the concn of BaCl2. (And note that Ba(OH)2 is quite soluble--I've never even looked for a Ksp for it).(Another note: look in your text and see if Ksp is listed for Ba(OH)2 in Ksp tables. If so we should use it. If not, we don't use it. :-)]
Assume not a Ksp problem.
Then Ba(OH)2 --> Ba^2+ + 2OH^-
If [Ba(OH)2] = 0.0062M then OH^- is twice that.
If you think it is a Ksp problem then use the BaCl2 as a common ion and solve for OH^-. Repost if you get stuck and explain the problem.
Woooo! the easy way was correct c: Thanks man. But i still kinda don't know why you just doubled it. I thought it would havesomething to do witht he ice table and such since it was in the common ion+buffer chapter :s
Care explaining or?
You want an ICE chart?
.........Ba(OH)2 ==> Ba^2+ + 2OH^-
initial..0.0062M......0........0
change..-0.0062......0.0062...2*0.0062
equil....0..........0.0062M..0.0134M
But I didn't do all of that in my mind. I looked at Ba(OH)2, I see there are 2 OH^- per molecule of Ba(OH)2 and the problem tells me the [Ba(OH)2] = 0.0062M so I know OH^- is twice that.
Wow im an idiot XD thanks.
And i don't want to be a nuisance, but how would i do the next question?
0.320M (NH4)2SO4 and 0.492M NH3
(Still looking for OH- conc)
Would i find the H+ concentration from the double dissociation of the NH4compound? (i know how to find the oh from there)
To calculate the concentration of OH- in the solution containing both Ba(OH)2 and BaCl2, you need to consider the dissociation of these compounds in water.
Let's start by calculating the concentration of Ba2+ ions contributed by BaCl2.
Given:
Concentration of BaCl2 = 1.00x10^(-2) M
BaCl2 dissociates into Ba2+ and 2Cl- ions. Since Ba2+ contributes to the OH- concentration through the subsequent reaction with OH-, we need to determine the concentration of Ba2+.
For every BaCl2 molecule, one Ba2+ ion is formed.
Hence, the concentration of Ba2+ ions contributed by BaCl2 is also 1.00x10^(-2) M.
Now, let's calculate the concentration of Ba2+ ions contributed by Ba(OH)2.
Given:
Concentration of Ba(OH)2 = 6.2x10^(-3) M
Ba(OH)2 dissociates into Ba2+ and 2OH- ions. Since we are interested in the OH- concentration, we can see that for every Ba(OH)2 molecule, two OH- ions are formed.
This means that the concentration of OH- ions contributed by Ba(OH)2 is twice the concentration of Ba(OH)2.
Concentration of OH- ions contributed by Ba(OH)2 = 2 * (6.2x10^(-3)) M = 1.24x10^(-2) M
Now, we have the concentrations of Ba2+ ions contributed by both BaCl2 and Ba(OH)2. We can use these concentrations to find the OH- ion concentration.
To do this, we need to consider the reaction between Ba2+ ions and OH- ions.
Ba2+ + 2OH- → Ba(OH)2
Using this balanced equation, we can see that for every Ba2+ ion, two OH- ions are consumed. Therefore, the concentration of OH- ions is equal to half the concentration of Ba2+ ions.
Concentration of OH- ions = (1.00x10^(-2) + 1.24x10^(-2)) / 2
= 1.12x10^(-2) M
So, the concentration of OH- ions in the solution containing 6.2x10^(-3) M Ba(OH)2 and 1.00x10^(-2) M BaCl2 is approximately 1.12x10^(-2) M.