Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.

0.16 M CH3NH3Cl

a mixture that is 0.16 M in CH3NH2
and 0.16 M in CH3NH3Cl

Answer 1

1). Calculating pH of the solution with 0.16 M CH3NH3Cl

Solution :-

Kb of CH3NH2 = 4.4*10^-4

Therefore ka of CH3NH3+ =Kw/Kb

                                            = 1*10^-14 / 4.4*10^-4

                                            = 2.27*10^-11

CH3NH3^+   + H2O ------ > CH3NH2 + H3O+

0.16 M                                          0                 0

-x                                                   +x              +x

0.16-x                                            x               x

Ka = [[CH3NH2][H3O+]/[CH3NH3+]

2.27*10^-11 = [x][x]/[0.16-x]

Since ka is very small therefore we can neglect the x from the denominator

Then we get

2.27*10^-11 = [x][x]/[0.16]

2.27*10^-11 * 0.16 = x^2

3.63*10^-12 = x^2

Taking square root of both sides we get

1.91*10^-6 =x

Therefore

pH= -log [H3O+]

pH= -log [1.91*10^-6]

pH= 5.72

2). a mixture that is 0.16 M in CH3NH2
and 0.16 M in CH3NH3Cl

Solution :- The mixture forms the buffer because weak base and its conjugate acids are mixed

Therefore using the Henderson equation we can calculate the pOH and then find pH

pOH= pkb + log [Acid]/[base]

pOH = (-log 4.4*10^-4)+ log [0.16/0.16]

pOH= 3.36 + 0

pOH = 3.36

pH +pOH = 14

therefore

pH = 14 – pOH

pH = 14 – 3.36

pH= 10.64