Question

Consider a titration involving 40.0 mL of 0.100 M ammonia , NH₃ (in an Erlenmeyer flask)with 0.100 M HCl (in a burette) to answer the next two questions.

NH₃ + HCl

Answer 1

1) answer : d) NH₃ and NH₄Cl

Explanation :

millimoles of HCl = 20 x 0.1 =2

millimoles of NH3 = 40 x 0.1 = 4

NH3 + HCl -------------------------> NH4Cl

4             2                                       0 --------------------------> initial

4-2          0                                        2   --------------------------> after reaction

2             0                                         2 ----------------------------> after reaction

in the mixture NH3 and NH4Cl are 2 millimoles remained

2) answer : b) Solve for the pH of an acidic salt using the K_a of NH₄Cl and its resulting concentration

Explanation :

NH3        +    HCl -------------------> NH4Cl

40 x 0.1         40 x0.1                   0       

4                     4                           0 -----------------------initial

0                     0                          4   ----------------------at equivalence point

here salt NH4Cl remained in the mixture

it is formed from the weak base and strong acid HCl. so it is acidic salt

salt concentration (C)= millimoles of salt/ total volume

                             = 4/(40+40)

                              = 0.05

pH of acidic salt in salt hydrolysis process: cation only partcipate in the reaction

NH4⁺ + H2O -----------------------> NH4OH + H⁺

0.05                                              0               0

0.05-x                                          x               x

Ka = [NH4OH][H⁺]/[NH4⁺]

Kb/Kw = (x)(x)/(0.05-x)

1.8 x 10^-5/ 1 x 10^-14 = x^2/(0.05-x)

by solving this we get x

x = [H+]

once we know [H+]

pH = -log [H+]

so correct answer is (b)