Show all work 10. (12 points) Tri ((CH3)3NH+Br , pha = 9.82) is a weak acid. You
ID: 589001 • Letter: S
Question
Show all work 10. (12 points) Tri ((CH3)3NH+Br , pha = 9.82) is a weak acid. You titrate 10.00 mL of 0.100 M trimethylammonium bromide with 0.100 M NaOH. Answer the following questions using activities. The activity coefficients are: methyl ammonium bromide trimethylammonium ion, 0.77; hydroxide ion, 0.76; hydronium ion, 0.83. (a) (3 points) Caleulate the equivalence point volume for the titration. (b) (3 points) What is the pH of the solution before any titrant is added? (c) (3 points) What is the pH when 8.52 mL of base has been added? (d) (3 points) What is the pH at the equivalence point?Explanation / Answer
a)
Vequivalence
mmol of waek acid = MV = 10*0.1 = 1
mmol of base required = 1
Vbase = mmol/V = 1/0.1 = 10 mL
Veq = Vbase = 10 mL
b)
pH before anything is added
First, assume the acid:
to be HA, for simplicity, so it will ionize as follows:
HA <-> H+ + A-
where, H+ is the proton and A- the conjugate base, HA is molecular acid
Ka = [H+][A-]/[HA]; by definition
initially
[H+] = 0
[A-] = 0
[HA] = M;
the change
initially
[H+] = + x
[A-] = + x
[HA] = - x
in equilbrirum
[H+] = 0 + x
[A-] = 0 + x
[HA] = M - x
substitute in Ka
Ka = [H+][A-]/[HA]
Ka = x*x/(M-x)
x^2 + Kax - M*Ka = 0
if M = 0.1 M; then
x^2 + (10^-9.82)x - 0.1 *(10^-9.82) = 0
solve for x
x =3.89*10^-6
substitute
[H+] = 3.89*10^-6
pH = -log(H+) = -log(3.89*10^-6) = 5.41
c)
mmol of acid = MV = 1
mmol ofbase = 0.1*8.52 = 0.852
after reaction
mmol of acid left = 1-0.852 = 0.148
mmol of conjguate base = 0.852
pH = pKa + log(A-/HA)
pH = 9.82 + log(0.852/0.148)
ph = 10.580
d
ph in equivalence
expect hydrolysis
since A- is formed
the next equilibrium is formed, the conjugate acid and water
A- + H2O <-> HA + OH-
The equilibrium s best described by Kb, the base constant
Kb by definition since it is an base:
Kb = [HA ][OH-]/[A-]
get ICE table:
Initially
[OH-] = 0
[HA] = 0
[A-] = M
the Change
[OH-] = + x
[HA] = + x
[A-] = - x
in Equilibrium
[OH-] = 0 + x
[HA] = 0 + x
[A-] = M - x
substitute in Kb expression
Kb = [HA ][OH-]/[A-]
Ka can be calculated as follows:
Kb = Kw/Ka = (10^-14)/(10^-9.82) = 6.61*10^-5
6.61*10^-5= x*x/(0.05-x)
solve for x
x^2 + Kb*x - M*Kb = 0
solve for x with quadratic equation
x = OH- =0.001785
[OH-] =0.001785
pOH = -log(OH-) = -log(0.001785) = 2.75
pH = 14-2.75= 11.25
pH = 11.25