I took 6.22 g of potassium hypochlorite and dissolved it up in a 100.0 mL flask.
I then added 8.00 mL of 2.00 M nitric acid.
The solution should be able to buffer that addition.
How much more (mLs) of the nitric acid can the system take before it is no longer in the buffer zone?
I took 6.22 g of potassium hypochlorite and dissolved it up in a 100.0 mL flask.
I then added 8.00 mL of 2.00 M nitric acid. The solution should be able to buffer that addition.
How much more (mLs) of the nitric acid can the system take before it is no longer in the buffer zone?
2 answers
mols KOCl = 6.22g/90.55 = 0.0687 or 68.7 millimols.
......OCl^- + H^+ ==> HOCl
I....68.7.....0........0
add..........16.............
C...-16.....-16........16
E....52.7.....0.........16
You can substitute to find the pH of this solution is 8.05
Out of the buffer range for pKa of 7.53 is 6.53.
Take the E line from above and add in x amount.
E ...52.7......0.......16
add............x.........
new C -x......-x.......+x
new E 52.7-x...0.......16+x
6.53 = 7.53 + log (52.7-x)/(16+x)
x = ? millimols HNO3. I think x is about 46 but you should confirm. Convert to mL of 2 M with M = millimols/mL
You should check it out. You added 16 mmols first, add to the 46 (or whatever number it is) to the 16. That gives OCl^- of 52.7-46 and HOCl of 16 + 46. Again, make sure these numbers are right and I noted that you must use 3 places to get the 6.53 pH.
Post your work if you get stuck.
......OCl^- + H^+ ==> HOCl
I....68.7.....0........0
add..........16.............
C...-16.....-16........16
E....52.7.....0.........16
You can substitute to find the pH of this solution is 8.05
Out of the buffer range for pKa of 7.53 is 6.53.
Take the E line from above and add in x amount.
E ...52.7......0.......16
add............x.........
new C -x......-x.......+x
new E 52.7-x...0.......16+x
6.53 = 7.53 + log (52.7-x)/(16+x)
x = ? millimols HNO3. I think x is about 46 but you should confirm. Convert to mL of 2 M with M = millimols/mL
You should check it out. You added 16 mmols first, add to the 46 (or whatever number it is) to the 16. That gives OCl^- of 52.7-46 and HOCl of 16 + 46. Again, make sure these numbers are right and I noted that you must use 3 places to get the 6.53 pH.
Post your work if you get stuck.
0 answers