Consider the titration of a 22.0 mL sample of 0.100 molL1 CH3COOH (Ka=1.8×105) w

Question

Consider the titration of a 22.0 mL sample of 0.100 molL1 CH3COOH (Ka=1.8×105) with 0.120 molL1 NaOH.

Determine each quantity:

Part A

the initial pH pH = 2.87

Part B

the volume of added base required to reach the equivalence point V = 18.3 mL

Part C

the pH at 5.0 mL of added base Express your answer using two decimal places. pH =

Part D

the pH at one-half of the equivalence point Express your answer using two decimal places. pH =

Part E

the pH at the equivalence point Express your answer using two decimal places. pH =

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 = 0.1 M; then

x^2 + (1.8*10^-5)x - 0.1*(1.8*10^-5) = 0

solve for x

x =0.00133

substitute

[H+] = 0 + 0.00133= 0.00133 M

[A-] = 0 + 0.00133 = 0.00133 M

[HA] = M - x = 0.1-0.00133= 0.09867 M

pH = -log(H+) = -log(0.00133 ) = 2.8761

b.

mmol of acid = MV = 22*0.1 = 2.2

mmol of base = MV = 18.3*0.12 = 2.193

after reaction

mmol ofa cid left = 2.20-2.193 = 0.007

mmol of conjugate formed = 2.193

pH = pKa + log(a-/HA)

pH = 4.75 + log(2.193/0.007)

pH = 7.25

c)

mmol of acid = MV = 22*0.1 = 2.2

mmol of base = MV = 5*0.12 = 0.6

after reaction

mmol ofa cid left = 2.20-0.6= 1.6

mmol of conjugate formed = 0.6

pH = pKa + log(a-/HA)

pH = 4.75 + log(0.6/1.6)

pH = 4.324

d)

in the half equivalenc epoint --> A- =HA

pH = pKa + log(a-/HA)

pH = 4.75

e)

in equivalence

A- + H2O <- >HA + OH-

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

[A-] = (22*0.1)/(22 + (22*0.1/0.12) = 0.054

(5.55*10^-10) = x*x/(0.054-x

x = sqrt(0.054 * (5.55*10^-10))

[OH-] = 0.00000547

pH = 14 + log(0.00000547) = 8.7379

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