Question

Calculate the pH for the titration of 40 mL of 0.5 M C5H5N with 0.2 M HCl at:a) 0 mL of titrant b) 5 mL before equivalence c) at equivalence d) 5 mL past equivalence

Kb for C5H5N = 1.7 X 10^-9

Answer 1

a)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

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

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

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

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (1.7*10^-9)x - (0.5)(1.7*10^-9) = 0

solve for x

x = 2.915*10^-5

substitute:

[OH-] = 0 + x = 2.915*10^-5

pOH = -log(2.915*10^-5) = 4.53

pH = 14 + pOH = 14 -4.53 = 9.47

b)

mmol of base = MV = 0.5*40 = 20

mmol of acid = MV = 5*0.2 = 1

after reaction

mmol of base left = 20-1 = 19

mmol fo conjgute formed = 1

pOH = pKb + log(Bh+/B)

pKb = -log(1.7*10^-9) = 8.77

pOH = 8.77 + log(1/19) = 7.491

pH = 14 - pOH = 14-7.491

pH = 6.509

c)

equivalence point

the next equilibrium is formed, the conjugate acid and water

BH+(aq) + H2O(l) <-> B(aq) + H3O+(aq)

The equilibrium is best described by Ka, the acid constant

Ka by definition since it is an base:

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

Ka can be calculated as follows:

Ka = Kw/Kb = (10^-14)/(4.4*10^-4) = 2.2727*10^-11

get ICE table:

Initially

[H3O+] = 0

[B] = 0

[BH+] = M

the Change

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = -x

in Equilibrium

[H3O+] = 0 + x

[B] = 0 + x

[BH+] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

(10^-14)/(1.7*10^-9) = x*x/(M-x)

V = (0.5*40)/(0.2) = 100 mL of acid erquired

Vtotal = 40+100 = 140

Mnew = 40*0.5/140 = 0.1428

5.88*10^-6 = x*x/(0.1428-x)

solve for x

x^2 + Ka*x - M*Ka= 0

solve for x with quadratic equation

x = H3O+ = 9.13*10^-4

[H3O+]  =9.13*10^-4

pH = -log([H3O+]) = -log(9.13*10^-4) = 3.039

pH = 3.039

d)

this is extra acid so

V = 100 + 5= 105 mmol ofa cid

mmol = MV = (0.2*5) = 1

[H+] = mmol/V = 1/105

pH = -log(0.009523) = 2.02