Posted by Andrew on Tuesday, June 12, 2012 at 12:58pm.
105 mL x 0.4M = 42.0 millimoles HOCl
105 mL x 0.36M = 37.8 mmoles NaOCl.
13.4 mL x 1.5M NaOH = 20.1 mmoles NaOH.
.........HOCl + OH^- ==> OCl^- + H2O
Initial..42.0...0........37.8.....
add............20.1.................
change..-20.1..-20.1.....+20.1
equil.....?.....0.........?
Substitute the equil line into the Henderson-Hasselbalch equation and solve for pH.
Related Questions
chemistry - A buffer solution was prepared by mixing 442 mL of 0.181 M NaOCl and...
Chemistry - I don't get this at all. A beaker with 105mL of an acetic acid ...
chemistry - what is the pH of a solution that is .15 M in HOCl and .25 M NaOCl ...
AP Chemistry - A 1.2M NaOCl solution is prepared by dissolving solid NaOCL in ...
apchemistry - If you have CL2 + H2o ---> H+ + Cl- + HOCl and you are ...
APChemistry - If you have CL2 + H2o ---> H+ + Cl- + HOCl and you are ...
APChemistry - If you have CL2 + H2o ---> H+ + Cl- + HOCl and you are ...
APChemistry - If you have CL2 + H2o ---> H+ + Cl- + HOCl and you are ...
chemistry - what is the [H3O+] of a solution that is .01 M in HOCl and .03 M in ...
chemistry - If you add 5.0 mL of 0.50 M NaOH solution to 20.0 mL to Buffer C, ...
For Further Reading
