Question

A 5.00ml volume of 0.20M solution contains 1.0mmol of solute. show the factor label calculation for why this statement is true.

Consider the titration of 0.1mmol of acetic acid in 25mL ( total volume) using 0.200M NaOH. assume Ka= 1.7 *10^-5

a) Calculate the total volume at the equivalence point.

b) calculate the acetate concentration at the equivalence point.

c) calculaye the theoretically expected pH at the equivalence point.

Please show work! Thanks!!!!

Answer 1

V = 5 ml

M = 0.2

m = 1 mmol of solute

this is true since:

m = M*V = 5ml*0.2 = 1 mmol

NOW

acetic acid

V = 25 ml

n = 0.1 mmol HA

M = 0.2 M of NaOH

M = ?

Ka = 1.7*10^-5

a)

total volume of equivalence point

in this point

moles of acetic acid = moles of base

0.1 mmol acetic acid = 0.1 mmol of base

0.1 mmol of base = M*V = V*0.2 M

M = mol/V

V = mol/V = 0.1/0.2 = 0.5 ml

Total volume V1+V2 = 25+.5 = 25.5 ml

b)

in the equivalent point

HA + NaOH --> NaA + H2O

1 mmol of HA + 1 mmol of base --> 1 mmol of NaA which is 1 mmol of A-

Therefore

[A-]= mmol/ml = 0.1 mmol / 25.5 ml = 0.003921 M

c)

theoreticall pH expected

this is hydrolysis

so expect this to form the next equilibrium

A- + H2O <--> HA + OH-

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

Kb can be calculated as follows:

Kb = Kw/Ka = (10^-14) / (1.7*10^-5) = 5.882*10^-10

in the euqilibrium

[HA] = [OH-] = x due to stoichiometry

[A-] = 0.003921 - x (considering the dissociated acid)

Susbtitue in Kb

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

5.882*10^-10 = (x*x) / (0.003921 - x)

since x is to small compared to 0.003921 (vs. 5.882*10^-10)

we can assume 0.003921 - x = 0.003921

5.882*10^-10 = (x*x) / (0.003921 )

x^2 = (5.882*10^-10)*0.003921 = 2.306*10^-12

x= sqrt( 2.306*10^-12) = 0.000001518

pOH = -log(OH) = -log(0.000001518) = 5.82

pH = 14- pOH = 14-5.82 = 8.18