Question

Consider the titration of 70.0 mL of 0.0300 M CH₃NH₂ (a weak base; K_b = 0.000440) with 0.100 M HNO₃. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 5.3 mL pH =	(c) 10.5 mL pH =
(d) 15.8 mL pH =	(e) 21.0 mL pH =	(f) 27.3 mL pH =

Answer 1

The trick to solve this kind of exercise is to know where you are in titration curve. It is well know that in the equivalence point: M_aV_a = M_bV_b

70 * 0.030 = 0.1 V_a

Va = 21 mL

So, at 21 mL is the equivalence point.

At 0 mL

CH₃NH₂+ H₂O ----------> CH₃NH₃⁺ + OH^-

i. 0.03 0 0

e. 0.03-x x x

Kb = x² / 0.03-x but Kb is a small value, so the value of x, therefore 0.03-x = 0.03

x = (4.40x10^-4 * 0.03)^1/2

x = [OH^-] = 3.63x10^-3 M

pOH = -log(3.63x10^-3)

pOH = 2.44

pH = 14-2.44 = 11.56

b) at 5.3 mL.

In this case, we have now the acid, so in order to calculate the pH, we need to use the HH equation which is:

pH = pKa + log [B]/[A] or pOH = pKb + log [S]/[B]

Writting now the equation:

CH₃NH₂+ HNO₃-------------> CH₃NH₃⁺ NO₃^-

If we now calculate the moles of HNO₃ and base:

moles A = 0.1 * 0.0053 = 0.00053 moles

moles B = 0.03 * 0.07 = 0.0021 moles

So, according to the above reaction, the remanent moles would be:

0.0021 - 0.00053 = 1.57x10^-3 moles

and the moles reacting would be 0.00053

and the concentration: 1.57x10^-3 / 0.0753 = 0.021 M

0.00053 / 0.0753 = 0.007 M

applying now the equation:

pOH = -log(4.40x10^-4) + log (0.007 / 0.021)

pOH = 2.88

pH = 14 - 2.88 = 11.12

c) at 10.5 mL

10.5 mL is actually the half equivalence, and at this point, pH = pKa, so:

Ka = Kw/Kb = 1x10^-14 / 4.40x10^-4 = 2.27x10^-11

pKa = 10.64 = pH

However, let's calculate this:

moles A = 0.1 * 0.0105 = 0.00105 moles reacting

moles remanent: 0.0021 - 0.00105 = 0.00105 moles

pOH = -log(4.40x10^-4) + log(0.00105/0.00105)

pOH = 3.36

14 - 3.36 = 10.64 = pH

d) at 15.8 mL

Same procedure as before so:

moles reacting = 0.1 * 0.0158 = 0.00158 moles

moles remanent = 0.0021 - 0.00158 = 0.00052 moles

pOH = 3.36 + log(0.00158/0.00052) = 3.84

pH = 14 - 3.84 = 10.16

e) 21 mL (equivalence point)

at this point all the base has been neutralized with the acid, so the moles reacting are the same for acid, which means:

moles of A = 0.0021 moles

the volume would be 91.5 mL or 0.091 L, so the concentration of the salt:

0.0021 / 0.091 = 0.023 M

now the reaction goes like this:

CH₃NH₃⁺ + H₂O ------------> CH₃NH₂ + H₃O⁺  Ka

i. 0.023 0 0

e. 0.023-x x x

Ka = x² / 0.023-x ----> but Ka is small too so, 0.023-x = 0.023

x = (0.023 * 2.27x10^-11)^1/2

x = [H₃O⁺] = 7.42x10^-7 M

pH = -log(7.24x10^-7)

pH = 6.14

f) 27.3 mL

At this point we already passed the equivalence point, All the base has been neutralised. With this large excess of a strong acid, the salt does not play any role in the pH of the final solution.

moles of A = 0.1 * 0.0273 = 0.00273 moles

moles of A remaining = 0.00273 - 0.0021 = 0.00063 moles

Concentration = 0.00063 / 0.0973 = 6.47x10^-3 M

pH = -log (6.47x10^-3) = 2.19

Hope this helps.