Question

A 0.2000M pyridine (Kb=1.7*10^-9) solution with initial volume 10.000ml is titrated with 2.500*10^-1 M HCL. what is the pH after the addition of :

a.) 2.00ml of HCL solution

b.) 6.00ml of HCL solution

c.)8.00ml of HCL solution

d.)10.00ml of HCL solution

Answer 1

For:

Pyridine ; let us assume Pyridine = B for simplicity

so

Pyridine + H2O <--> (Pyridine H+ + OH- becomes:

B + H2O <-> HB+ + OH-

Kb = [HB+][OH-]/[B]

so....

a)

2 mL of Hcl

mmol of acid added = MV = (2.5*10^-1)*2 = 0.5 mmol of acid

this reacts with base:

mmol of base = MV = 0.2*10= 2 mmol of base

mmol of base after reaction = 2.0 -0.5 = 1.5 mmol of base

and conjugate forms:

mmol of conjugate formed = 0 + 0.5 = 0.50

this becomes a buffer, sicne there is weak base+ conjugate acid

pOH = pKb + log(BH+/B)

pKB = -log(Kb) = -log(1.7*10^-9) = 8.7695

so

pOH = 8.7695+ log(0.50/1.5)

pOH = 8.292

pH = 14-pOH = 14-8.292= 5.708

c)

V = 6 mL of acid

mmol of acid = MV = 0.25*6= 1.5 mmol of acid

mmol of base after reaction = 2-1.5 = 0.5

mmol of conjugate formed = 0 + 1.5= 1.5

this is also a buffer... special one , since base = conjugate acid

so

pOH = pKb + log(BH+/B)

pOH = 8.292+ log(0.5/1.5) = 7.814878

pH = 14-7.814878= 6.1851

c)

V = 8 mL of acid

mmol of acid = MV = 0.25*8 =2

this is equivalence point!

mmol of acid = mmol of base

so..

all base/acid reacts

Vtotal = V1+V2 = 10+8= 18mL of total solution

calculate conjugate formed

mmol of conjguate formed = MV = 8*0.25 = 2.0 mmol of conjugate

recalcualte concentration of conjugate

[BH+] = mmol/Vt = 2.0/18= 0.11111 M of BH+

so..

BH+ + H2O <-> H3O+ + B

so

Ka = [H3O+][B]/[BH+]

Ka = Kw/Kb = (10^-14)/(1.7*10^-9) = 5.882*10^-6

so

[H3O+] = x = [B]

[BH+] = M-x = 0.11111 -x

substitute

Ka = [H3O+][B]/[BH+]

becomes

5.882*10^-6 = x*x/(0.11111 -x)

solve for x

x = 5.29*10^-4

[OH-] = x = 5.29*10^-4

pOH = -log(OH-) -log(5.29*10^-4 ) = 3.2765

pH = 14-pOH = 14-3.2765= 10.7235

f)

finally...

mmol of acid added = MV = 10*0.25*0.1 = 2.5 mmol of acid

mmol of base = 2.0

mmol of acid left = 2.5-2. = 0.5 mmol of acid left

Vtotal = 10+10 = 20 mL

[acid] = 0.5/20= 0.025 of acid

pH = -log(H+) = -log(0.025 )

pH = 1.6020