A solution is prepared by mixing 100 mL of 0.500 M NH3 and 100 mL of 0.500 M HCl. Assuming that the volumes are additive, what is the pH of the resulting mixture?

Question

A solution is prepared by mixing 100 mL of 0.500 M NH3 and 100 mL of 0.500 M HCl. Assuming that the volumes are additive, what is the pH of the resulting mixture?
kb for ammonia is 1.8x10-5

DrBob222 · Answer

The base and acid are exactly neutralized; the pH is determined solely by the hydrolysis of the salt.
(NH4Cl) = 50 mmols/200 mL = 0.25M.

......NH4^+ + H2O..> NH3 + H3O^+
I.....0.25.............0......0
C......-x..............x......x
E.....0.25-x...........x......x

Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.25-x)
Solve for x = (H3O^+) and convert that to pH.