Question

9. You have prepared a solution 0.2 M ammonium chloride (NH4Cl). Calculate the pH of this solution when the compounds below have been added. The pKa of NH4 + is 9.25.

a. 0.025 M NaCl

b. 0.01 M NaOH

c. 0.05 M NaOH

d. 0.25 M NaOH

Answer 1

a ) p^H= p^ka + log [ salt ] / [ Acid ]

= 9.25 + log [ 0.025 ] / [ 0.2 ]

= 9.25 + log 0.125

p^H =14 - 8.34= 5.66

b) NH₄Cl + NaOH-------------------> NH₃ + NaCl + H₂O

[ NH₄C ] = 0.2 + 0.01 = 0.21 and [ NH₃ ] = 0.2 - 0.01 = 0.19

p^H = 14 - 9.25 + log [ 0.21 ] / [ 0.19 ] = 14 - 9.25 + log 1.10 =3.10

Change in pH = = 5.66 - 4.69 = PH increase = 0.97

c) NH₄Cl + NaOH-------------------> NH₃ + NaCl + H₂O

[ NH₄C ] = 0.2 + 0.05 = 0.21 and [ NH₃ ] = 0.2 - 0.05 = 0.19

p^H = 14 - 9.25 + log [ 0.25 ] / [ 0.15 ] = 14 - 9.25 + log -1.52 =3.23

Change in pH = = 5.66 - 3.23 = PH increase = 2.43 increses

d) NH₄Cl + NaOH-------------------> NH₃ + NaCl + H₂O

[ NH₄C ] = 0.2 + 0.25 = 0.45 and [ NH₃ ] = 0.2 - 0.25 =0.05

p^H = 14 - 9.25 + log [ 0.45] / [ 0.05] = 14 - 9.25 + log 9 =3.8

Change in pH = = 5.66 - 3.8 = PH increase = 1.86