In: Chemistry
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
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