Consider the titration of 30.0 ml of 0.050 m nh3 with 0.025 mhcl. calculate the ph after the following volumes of titrant have been added.

a) 0 mL HCl added:

⇒ pH = 10.973

b) 20mL HCl added:

⇒ pH = 12

c) 70 mL HCl added:

⇒ pH = 2.602

Explanation:

NH3 ↔ NH4+  + OH-

∴ Kb = 1.8 E-5 = [OH-][NH4+]/[NH3]

∴ C NH3 = 0.050 m = (0.050 mol/Kg) (1000Kg/m³)(m³/1000L) = 0.050 mol/L

equivalent point:

∴ (VNH3)(C NH3) = (VHCl)(C HCl)

⇒ (0.03 L)(0.050 mol/L) = (VHCl)(0.025 mol/L)

⇒ V HCl = 0.06 L

0 mL HCl added:

mass balance:

⇒ C NH3 = [NH3] + [NH4+] = 0.050

charge balance:

⇒ [OH-] = [NH4+]

⇒ 0.050 = [NH3] + [OH-]

⇒ [NH3] = 0.050 - [OH-]

⇒ Kb = [OH-]² / (0.050 - [OH-]) = 1.8 E-5

⇒ [OH-]² + 1.8 E-5[OH-] - 9 E-7 = 0

⇒ [OH-] = 9.397 E-4 mol/L

⇒ pOH = - Log [OH-]

⇒ pOH = 3.027

⇒ pH = 10.973

after 20 mL HCl added:

⇒ C NH3 = ((0.03)(0.050)) - ((0.02)(0.025)))/(0.03+0.02) = 0.02 mol/L

⇒ C HCl =  (0.02)(0.025)/(0.03+0.02) = 0.01 mol/L

mass balance:

⇒ 0.02 + 0.01 = [NH3] + [NH4+] = 0.03

charge balance:

⇒ [NH4+] + [H3O+] = [OH-]

∴ [H3O+] = C HCl = 0.01 mol/L

⇒ [NH4+] = [OH-] - 0.01

⇒ [NH3] = 0.03 - ( [OH-] - 0.01 )

⇒ [NH3] = 0.04 - [OH-]

⇒ Kb = ([OH-]([OH-] - 0.01))/(0.04 - [OH-]) = 1.8 E-5

⇒ (1.8 E-5)(0.04 - [OH-]) = [OH-]² - 0.01[OH-]

⇒ 7.2 E-7 - 1.8E-5[OH-] = [OH-]² - 0.01[OH-]

⇒ [OH-]² - 9.982 E-3[OH-] - 7.2 E-7 = 0

⇒ [OH-] = 0.020 mol/L

⇒ pOH = 1.696

⇒ pH = 12.30

c) after 70 mL HCl added:

⇒ C HCl = ((0.07)(0.025))-((0.03)(0.050))/(0.07+0.03) = 2.5 E-3 mol/L

⇒ [HCl] = C HCl = 2.5 E-3 mol/L

⇒ pH = - Log [HCl] = 2.602