- #1

Mitchtwitchita

- 190

- 0

NH4+ ---> NH3 + H+

0.36 0.30 0

-x +x +x

0.36-x 0.30 + x x

Ka=[NH3][H+]/[NH4+]

5.6 x 10^-10 = (0.30 + x)(x)/(0.36 - x)

5.6 x 10^-10 = 0.30x/0.36, assuming x is negligible

=6.72 x 10^-10

pH=-log[H+]

=-log (6.72 x 10^-10)

=9.17

I don't know what to do next, can anybody please point me in the right direction?