Question

Consider the titration of 30.0 mL of 0.030 M NH₃ with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added.

(a) 35.0 mL

(b) 36.0 mL

(c) 37.0 mL

Answer 1

A) Moles NH₃ = 0.0300 L x 0.030 M=0.00090
moles HCl = 0.035 L x 0.025 M= 0.00087

NH₃ + H⁺ ----> NH₄⁺

moles of NH₃ left = 0.00090 - 0.00087 =0.00003
moles of NH₄⁺ formed = 0.00087
total volume = 30.0 + 35.0 = 66.0 mL => 0.065 L
[NH₃]= 0.00003 / 0.065 = 4.61 x 10^-4 M
[NH₄⁺]= 0.00087/ 0.067 = 0.0129 M

pKb = - log Kb = 4.7

pOH = pK_b + log salt/base
pOH = 4.7 + log 0.0129 / 4.61 x 10^-4 = 6.14
pH = 14 - 6.14 = 7.85

B) Moles NH₃ = 0.0300 L x 0.030 M=0.00090
moles HCl = 0.036 L x 0.025 M= 0.0009

NH₃ + H⁺ ----> NH₄⁺

moles of NH₃ left = 0.00090 - 0.0009 =0
moles of NH₄⁺ formed = 0.0009
total volume = 30.0 + 36.0 = 66.0 mL => 0.066 L
[NH₃]= 0
[NH₄⁺]= 0.0009/ 0.067 = 0.0134 M

K_a = K_w / K_b

K_a = 10^-14/1.8 x 10^-5 = 5.56 x 10^-10

                     NH₄⁺       H₂O             NH₃       H₃O⁺

initial    0.0134                            -    0    0

change                      -x                            -                         +x                                  +x

equilibrium            0.0134-x                             -                               x                                    x

K_a = [NH₃][H₃O⁺]/[NH₄⁺]

1.8 x 10^-10 = x²/0.0134      (assume x<<0.0134)

x = 0.155 x 10^-5 M [H⁺]

pH = -log[H⁺]

pH = -log(0.155 x 10^-5)

pH = 4.80

C) Moles NH₃ = 0.0300 L x 0.030 M=0.000900
moles HCl = 0.037 L x 0.025 M= 0.000925

NH₃ + H⁺ ----> NH₄⁺

moles of HCl left = 0.000925 - 0.000900 =0.000025
moles of NH₄⁺ formed = 0.0009
total volume = 30.0 + 37.0 = 67.0 mL => 0.067 L
[HCl]= 0.000025 / 0.067 = 3.73 x 10^-4 M
[NH₄⁺]= 0.0009/ 0.067 = 0.0134 M

pK_a = - log K_a = 9.74

pH = pK_a + log salt/acid
pH = 9.74 + log 0.0134 / 3.73 x 10^-4 = 11.29