Determine the PH of a solution that results from mixing 50 ml of .200 M hydroazoic acid, HN3 with 50 ml of CA(OH)2
3 answers
You don't need M Ca(OH)2??? I don't think you can work the problem without it. .
sorry that's .2 M caoh2
You meant Ca(OH)2.
This is new to me but I believe this is an acid/base reaction between HN3 and Ca(OH)2. I think the equation is
2HN3 + Ca(OH)2 ==> Ca(N3)2 + 2H2O
mols HN3 = M x L = 0.2 x 0.05 = 0.01
mols Ca(OH)2 = M x L = 0.2 x 0.05 = 0.01
We start with this probably being a limiting reagent problem. I work those the long way. First, how much Ca(N3)2 could be formed from 0.01 mol Ca(OH)2? That will be 0.01 x [1 mol Ca(N3)2/1 mol Ca(OH)2] = 0.01 mol Ca(N3)2
How much Ca(N3)2 could be formed from 0.01 mol HN3? That's 0.01 mol HN3 x (1 mol Ca(N3)2/2 mol HN3) = 0.01 x 1/2 = 0.005 mols Ca(N3)2; therefore, HN3 is the limiting reagent and you will have 0.005 mols of the salt formed, all of the HN3 will be used and you will have 0.005 mols Ca(OH)2 in excess. Convert that to M = 0.005 mols/0.1L = 0.05M
Now all you need to do is to convert 0.05M Ca(OH)2 to pH. The HN3 has a Ka of 1.9E-5; therefore, it will contribute negligible OH or H.
This is new to me but I believe this is an acid/base reaction between HN3 and Ca(OH)2. I think the equation is
2HN3 + Ca(OH)2 ==> Ca(N3)2 + 2H2O
mols HN3 = M x L = 0.2 x 0.05 = 0.01
mols Ca(OH)2 = M x L = 0.2 x 0.05 = 0.01
We start with this probably being a limiting reagent problem. I work those the long way. First, how much Ca(N3)2 could be formed from 0.01 mol Ca(OH)2? That will be 0.01 x [1 mol Ca(N3)2/1 mol Ca(OH)2] = 0.01 mol Ca(N3)2
How much Ca(N3)2 could be formed from 0.01 mol HN3? That's 0.01 mol HN3 x (1 mol Ca(N3)2/2 mol HN3) = 0.01 x 1/2 = 0.005 mols Ca(N3)2; therefore, HN3 is the limiting reagent and you will have 0.005 mols of the salt formed, all of the HN3 will be used and you will have 0.005 mols Ca(OH)2 in excess. Convert that to M = 0.005 mols/0.1L = 0.05M
Now all you need to do is to convert 0.05M Ca(OH)2 to pH. The HN3 has a Ka of 1.9E-5; therefore, it will contribute negligible OH or H.