Question

What is the pH of a 4.30×10-2 M aqueous solution of ammonium nitrate, NH4NO3 ?

Answer 1

00120 Concentration of NH, = 0.0430M ICE table NH, + H2O + NH, + H2O I(M): 0.0430 0 0 C(M): -X +x +x E(M): 0.0430 - x Baseionization constant of NH3,K, = 1.8x10- 1.0x10-14 Acidionization constant of NH, 1.8x10-5-5.6x10-10 Acidionization constant, K = [NH,] (x)(x) (0.0430- x) 5.6x10-10 = .82 5.6x10-10 = 2 (0.0430-x) Assume x is much smaller than 0.0430, so (0.0430-x)= 0.0430 XP 5.6 x 10-10 - 0.0430 x = (5.6x10-10) (0.0430) = 24.1x10-12 x= 24.1*10-12 = 4.91x10-6 M According to the equilibrium table, H,0* 7 = x = 4.91x10-6 M pH = -log[H.0*] = -log(4.91x106 M) = 6-log(4.91) =5.31 Therefore , pH of the solution is 5.31