Consider the titration of a 21.0 mL sample of 0.110 molL1 CH3COOH ( K a=1.8×105)
ID: 590929 • Letter: C
Question
Consider the titration of a 21.0 mL sample of 0.110 molL1 CH3COOH (Ka=1.8×105) with 0.125 molL1 NaOH. Determine each quantity:
a) the initial pH. Express your answer using two decimal places.
b) the volume of added base required to reach the equivalence point
c) the pH at 4.0 mL of added base. Express your answer using two decimal places.
d) the pH at one-half of the equivalence point. Express your answer using two decimal places.
e) the pH at the equivalence point. Express your answer using two decimal places.
f) the pH after adding 6.00mL of base beyond the equivalence point. Express your answer using two decimal places.
Explanation / Answer
a)
First, assume the acid:
CH3COOH
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 = 0.11 M; then
x^2 + (1.8*10^-5)x - 0.11 *(1.8*10^-5) = 0
solve for x
x =0.01434
substitute
[H+] = 0 + 0.01434= 0.01434 M
pH = -log(0.01434) = 1.834
b)
V for equivalence
mmol of acid = MV = 21*0.11 = 2.31
Vbase = mmol/M = 2.31/0.125 = 18.48 mL
c)
mmol of acid = 21*0.11 = 2.31
mmol of base = MV = 0.125*4 ?= 0.5
after reaction
mmol of acid = 2.31-0.5 = 1.81
mmol of acetate = 0.5
pH = 4.75 + log(0.5/1.81) = 4.19
d)
in the 50% point
pH = pKa + log(50/50)
pH= 4.75 + log(1)
pH = 4.75
e
equivalence
Vtotal = 18.48+21 = 39.48 mL
[A-] = 2.31/39.48 = 0.05851 M
expect hydrolysis
Let HA --> CH2OH and A- = CH2O- for simplicity
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-]
5.55*10^-10 = x*x/(0.05851 -x)
solve for x
x^2 + Kb*x - M*Kb = 0
solve for x with quadratic equation
x = OH- =5.698*10^-6
[OH-] =5.698*10^-6
pOH = -log(OH-) = -log(5.698*10^-6) = 5.24
pH = 14-5.24= 8.76
f)
extra base:
V of base = (18.38+6) = 24.38
Vtotal = 21+24.38 = 45.38
[OH-] = (6)(0.125) / 45.38
[OH-]= 0.016527
pH = 14+ log(0.016527) = 12.22