At 25 degrees Celsius, the base ionization constant for NH3 is 1.8 x ^-5. Determine the percentage ionization of a M solution of ammonia at 25 degree Celsius.
3 answers
You need to insert the molarity.
Sorry, its 0.150 M solution of ammonia at 25 degrees celsius
...........NH3 + H2O ==> NH4^+ + OH^-
initial..0.150M...........0........0
change.....-x.............x........x
equil...0.150-x............x.......x
Kb = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
Substitute into Kb expression and solve for x = (OH^-).
Then %ion = [(OH^-)/0.1]*100 = ?
initial..0.150M...........0........0
change.....-x.............x........x
equil...0.150-x............x.......x
Kb = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
Substitute into Kb expression and solve for x = (OH^-).
Then %ion = [(OH^-)/0.1]*100 = ?
1 answer