- #1
rokku
- 7
- 0
What amount of sodium Nitrate must be added to 500 ml of .200 M solution of HN02. Ka=4.0^-4
Ka=(H+)(A-)
--------
(HA)
Ph of 5 = 1.0^-5 (H+) so, (1.0^-5)(x)
---------- = 4.0^-4 , however X turns out to be way too big
.200
Ka=(H+)(A-)
--------
(HA)
Ph of 5 = 1.0^-5 (H+) so, (1.0^-5)(x)
---------- = 4.0^-4 , however X turns out to be way too big
.200