15. Consider the titration of a 25.00 ml sample of 0.125 M (CH3)2NH with 0.150 M
ID: 577205 • Letter: 1
Question
15. Consider the titration of a 25.00 ml sample of 0.125 M (CH3)2NH with 0.150 M HClO4. Determine each of the following. All work must be shown for credit. (a) the pH before addition of HClO4. (b) the volume of HClO4 required to reach the equivalence point. (c) the pH after addition of 5.0 ml of HClO4. (d) the pH at half the equivalence point. (e) the pH after adding 5.00 ml of HClO4 beyond the equivalence point. (f) Neatly sketch the titration curve. Clearly label both axes of the titration curve. Label the equivalence point and the half equivalence point. Identify the buffer region.Explanation / Answer
For:
(CH3)3NH; let us assume (CH3)3NH= B for simplicity
so
(CH3)3NH+ H2O <--> (CH3)3NH2+ + OH- becomes:
B + H2O <-> HB+ + OH-
Kb = [HB+][OH-]/[B]
so....
a)
no volume of acid
B + H2O <-> HB+ + OH-
Kb = [HB+][OH-]/[B]
in equilibrium
for 1 mol of OH- we hava always 1 mol of HB+
[HB+] = [OH-] = x (the dissociation fraction of acid)
[B] = M-x = 0.07-x (Account for dissociation!)
pKb = 3.16
(10^-3.16) = x*x/(0.125-x)
solve with quadratic formula
x = [OH-] = 0.008959
pOH = -log(Oh-) = -log(0.008959) = 2.05
pH = 14-pOH = 14-2.05= 11.95
b)
Volume erquired
mmol of base = MV = 0.125*25 = 3.125 mmol
Vacid = mmol/V = 3.125/0.150 = 20.83 mL
c)
5 mL of acid:
mmol of acid added = MV = 0.15*5 = 0.75 mmol of acid
this reacts with base:
mmol of base = MV = 25*0.125 = 3.125 mmol of base
mmol of base after reaction = 3.125 -0.75 = 2.375 mmol of base
and conjugate forms:
mmol of conjugate formed = 0 + 0.53 = 0.53
this becomes a buffer, sicne there is weak base+ conjugate acid
pOH = pKb + log(BH+/B)
pOH = 3.16+ log(0.75 /2.375 )
pOH = 2.66
pH = 14-pOH = 14-2.66= 11.34
d)
in half equivalence point
mmol of base = mmol of conjguate acid
B = HB+
pOH = pKb + log(BH+/B)
pOH = pKb
pOH = 3.16
pH = 14-3.16 = 10.84
d)
V = 20.83 mL of acid
Vtotal = 25+20.83 = 45.83
this is equivalence point!
mmol of acid = mmol of base
so..
all base/acid reacts
calculate conjugate formed
mmol of conjguate formed = MV = 25*0.125 = 3.125 mmol of conjugate
recalcualte concentration of conjugate
[BH+] = mmol/Vt = 3.125 / 45.83 = 0.068186
so..
BH+ + H2O <-> H3O+ + B
so
Ka = [H3O+][B]/[BH+]
Ka = Kw/Kb = (10^-14)/(10^-6.16) = 1.445*10^-8
so
[H3O+] = x = [B]
[BH+] = M-x = 0.068186-x
substitute
Ka = [H3O+][B]/[BH+]
becomes
1.445*10^-8= x*x/(0.068186-x)
solve for x
x = 3.138*10^-5
[OH-] = x = 3.138*10^-5
pOH = -log(OH-) -log(3.138*10^-5 ) = 4.50
pH = 14-pOH = 14-4.50 = 9.5
e)
excess
[H+] = mmol/V = (5*0.125)/(25+25) = 0.0125
pH = -log(0.0125) = 1.903
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.