Question

a 15.0 ml sample of 0.250 M NH3 is titrated with 0.500 M HCL (aq). what is the ph if we add 10.0 ml of the acid?

Answer 1

The reactions involved are as follows:

1) NH₃(aq) + HCl(aq) NH⁴⁺(aq) + Cl^-(aq)

2) NH⁴⁺(aq) + H₂O(l) NH₃(aq) + H₃O⁺(aq)

1st calculates the limiting reagent in the reaction 1. must calculate the moles of NH₃ and HCl

The moles of NH₃ and HCl are calculated according to the volume and concentration of the molarity formula:

M = mol/l     mol = Mxl = 0.250 (mol/l)x0.015 (l) = 0.00375 mol NH₃

M = mol/l     mol = Mxl = 0.500 (mol/l)x0.010 (l) = 0.00500 mol HCl

The limiting reagent is NH₃, so the moles of NH₃ and NH₄⁺ are equal to (1) = 0.00375 moles.

In reaction (2) is proposed for calculating the balance. The equilibrium constant of NH₃ Kc = 1.8x10^-5.

K_c = [NH₃][H₃O⁺] / [NH₄⁺]

K_c = [X][X] / [0.00375 - X]

K_c = X²/(0.00375 - X)

0.00375 K_c - (X)(K_c) = X²

(0.00375)(1.8x10^-5) - (1.8x10^-5X) = X²

(6.75X10^-8) = X² + 1.8X10^-5X

X² + 1.8X10^-5X - 6.75X10^-8 = 0

2nd grade equation is solved and you get the concentration of protons.

X = 2.51x10^-4

It calculates pH: pH = -log [H₃O⁺]

pH = -log (2.51x10^-4)

pH = 1.8