Calculate the [NH4+] in solution that is 0.25 M NH3 and 0.20 M NaOH. (Kb for NH3 = 1.8 x 10-5)
I set up ICE from the equation:
NH3 + H20 ------NH4 + OH-
I 0.25 0 0
C -x +x +x
E 0.25-x x x
1.8 *10^-5 = x*x/ 0.25-x
4.5* 10^-6 = x2 took the square root to get:
2.12 * 10^-3 =x
but I got it incorrect so I am wondering what I did wrong in my set up.
2 answers
Where did you factor in NaOH? The ice chart looks ok. But you must remember that OH^- = x from the NH3 + H2O reaction and 0.2 M from NaOH; therefore, the total OH^- = x+0.2.
Thanks I figured out what I didn't do which was factor in that 0.2M of NAOH
2 answers