What concentration of of CN- is needed to produce the same POH as 3.5*10^-5M solution of BA(OH)2?
.........Ba(OH)2 ==> Ba^2+ + 2OH^-
I.......3.5E-4........0.......0
C......-3.5E-4.....3.5E-4..2*3.5E-4
E........0.........3.5E-4...7.0E-4
So the (OH^-) from Ba(OH)2 = 7.0E-4. We want to know how much CN^- it will take to produce OH^- = 7.0E-4 M.
.........CN^- + HOH ==> HCN + OH^-
I........x...............0.....0
C......-7.0E-4........7.0E-4..7.0E-4
E.....x-7.0E-4........7.0E-4..7.0E-4
Kb for CN^- = (Kw/Ka for HCN) = (7.0E-4)^2/(x-7.0E-4)
Solve for x = molar concn of CN^- required. Kw is 1E-14. You will look up Ka for HCN.
I got 2*10^-5 for the Kb, is that right?
When you ask a question like this you should tell us what you used for Ka. Our tables may not give us the same value as your table(s). Therefore, we don't know if you've done it right or not. I used 6.2E-10 for Ka for HCN and using that value Kb for CN^- is 1.6E-5.
To find the concentration of CN-, we need to compare it to the concentration of BA(OH)2 using the concept of stoichiometry and the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction between CN- and BA(OH)2:
2 CN- + BA(OH)2 -> BA(CN)2 + 2 OH-
From the equation, we can see that 2 moles of CN- react with 1 mole of BA(OH)2 to produce 2 moles of OH-.
Next, we need to find the concentration of OH- in the 3.5*10^-5M solution of BA(OH)2. Since the molarity of BA(OH)2 is 3.5*10^-5M, the concentration of OH- is also 3.5*10^-5M.
Now, we can use the ratio from the balanced equation to find the concentration of CN-.
Since 2 moles of OH- are produced for every 2 moles of CN- reacting with 1 mole of BA(OH)2, the concentration of CN- will be the same as the concentration of OH-. Therefore, the concentration of CN- needed to produce the same POH as the 3.5*10^-5M solution of BA(OH)2 is also 3.5*10^-5M.