Calculate the pH of 0.5 moldm-3 ammonium chloride at 25C?
Ka = 5.6 x 10^-10 concentration is 0.50 m
Answers
Old Science Guy
... no, it's 0.50 M not m NH4+1 + H2O ↔ NH3 + H30+ Kh = [NH3] [H3O+] / [NH4+] [NH3] = [H3O+] and Kh = Kw/Ka so 1E-14/5.6E-10 = X^2 / (0.50-X) but X is very small so 0.50-X is essentially 0.50 1.7857E-5 = X^2 / 0.5 X^2 = 8.9286E-6 X = 2.99E-3 pH = -log(2.99E-3) = 2.52 which is two s.f. When you get a good response, please consider giving a best answer. This is the only reward we get.
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