Answer to Question #118590 in General Chemistry for Natasha Luis Vera

Question

For the titration of 20.00 mL of 0.1206 M NH3 with 0.0947 M HCl, calculate the pH before the addition of any titrant, at 10.00 mL prior to equivalence point, at the equivalence point, and at 10.00 mL after the equivalence point. (Kb (NH3) = 1.76 x 10-5 )

Answer 1

The reaction between ammonia and hydrochloric acid is the following

NH₃*H₂O+ HCl = NH₄Cl + H₂O

Initially, without addition of hydrochloric acid, the solution contains only ammonia.

The reaction of ammonia dissociation is the following

NH₃*H₂O <-> NH₄⁺ + OH^- with K_b = [NH₄⁺]*[OH^-]/[NH₃*H₂O] = 1.76*10^-5

[NH₄⁺] = [OH^-] -> K_b = [OH^-]²/[NH₃*H₂O];

[OH^-] = K_w/[H⁺] -> K_b = (K_w/[H⁺])²/[NH₃*H₂O] -> [H⁺] = K_w/(K_b*[NH₃*H₂O])^0.5 = 6.87*10^-12 mol/L

pH = 11.16

In the equivalent point c(NH₃*H₂O)*V(NH₃*H₂O) = c(HCl)*V(HCl)

V(HCl) = 0.1206 mol/L*20 ml/0.0947 mol/L = 25.47 mL

At 10 ml prior to equivalence point (at 15.47 ml of added HCl solution) pH value of the solution is calculated as

pH = 14 - pK_b + lg([NH₃*H₂O]/[NH₄Cl]) = 14 - 4.75 - lg(([NH₃*H₂O]_initial-[NH₃*H₂O]_titrated)/[HCl]_added) = 9.06

In the equivalence point

Ka = Kw/Kb

Let x mol/L of NH₄⁺ be formed during neutralization of ammonia by HCl, x mol/L of NH₃ will react and (0.1206 mol/L*20 ml/(20 ml + 25.47 ml)) = 0.053 mol/L is the initial amount of the NH₃ in the solution. After the reaction with HCl it will remain (0.053 -x) mol/L of NH₃

Then the balance between NH₃ and NH₄⁺ will be written using Ka (Ka = Kw/Kb)

In accordance with Ka = [NH₃]*[H⁺]/[NH₄⁺]

x²/(0.053-x) = 10^-14/(1.76*10^-5)

x = 5.49*10^-6 mol/L

pH = -lgx = 5.26 in the equivalence point

After adding extra 10 ml of HCl (25.47+10 = 35.47 ml) after equivalence point

pH value is determined by the amount of HCl added

c(H⁺) = (10 ml*0.0947 mol/L)/35.47 ml = 0.0267 mol/L

pH = -lg(c(H⁺)) = 1.57