Question

Weak diprotic acid H₂A is titrated with 0.2000 M NaOH. The initial concentration of H₂A is 0.1000 M and the initial volume is 50.00 mL.      K_a1 = 1.0 x 10⁻⁵       K_a2 = 1.0 x 10⁻⁹

(a) (0.2 pt) Identify the predominate species at the first equivalence point, by chemical formula:

(b) (0.5 pts) Calculate the pH of the acid solution before any base is added.

(c) (0.5 pts) Calculate the pH at the first equivalence point.

(d) (0.2 pts) How many mL base, total, must be added to achieve the 2nd equivalence point?

(e) (0.5 pts) Calculate the pH at the second equivalence point.

Answer 1

Solution :-

(a) (0.2 pt) Identify the predominate species at the first equivalence point, by chemical formula:

Solution :- H2A + NaOH ----- > HA^- + H2O

so the predominant species at the first equivalence point HA^-

(b) (0.5 pts) Calculate the pH of the acid solution before any base is added.

Solution :-

ka1= [HA-][H3O+]/[HA]

1.0*10^-5 [x][x]/[0.100-x]

since ka is small we can neglect the x from the denominator

1.0*10^-5 *0.100 = x^2

1*10^-6 = x^2

taking square root of both sides we get

1.0*10^-3 = x=[H3O+]

pH= -log [H3O+]

pH= -log [1.0*10^-3]

pH= 3

(c) (0.5 pts) Calculate the pH at the first equivalence point.

Solution :- At the first equivalence point all acid is converted into the HA^-

so the pH at the first equivalence point is same as pka1

pka1 = -log [ka1]

pka1 = -log [1.0*10^-5]

pka1= 5

so the pH at the first equivalence point = 5

(d) (0.2 pts) How many mL base, total, must be added to achieve the 2nd equivalence point?

Solution :-

moles of acid = molarity * volume in liter

                    = 0.100 mol per L * 0.050 L

                    = 0.005 mol

mole ratio of the acid to base is 1 : 2

threfore moles of base needed = 0.005*2 = 0.01 mol base

now lets calculate the volume of the base

volume in liter = moles / molarity

                    = 0.01/0.200 mol per L

                    = 0.05 L

0.05 L * 1000 ml / 1 L = 50.0 ml

so volume of base needed = 50.0 ml

(e) (0.5 pts) Calculate the pH at the second equivalence point.

at the second equivalence point all HA^- is converted to A^2-

so the need to use the kb2 to calculagte the pH

new molarity of the A^2- at total volume of 0.100 L = 0.005/0.1 L = 0.05 M

kb2 = kw/ ka2

kb2= 1*10^-14 / 1.0*10^-9 = 1*10^-5

kb2= [AH^-][OH^-]/[A^2-]

kb2 = [x][x]/[A^2-]

1*10^-5 = x^2 /0.05

1*10^-5 *0.05 =x^2

5*10^-7 =x^2

taking square root of both side we get

7.07*10^- M= x = [OH-]

pOH= -log [OH-]

pOH = -log [7.07*10^-4]

pOH =3.15

pH + pOH = 14

pH= 14 - pOH

pH= 14 - 3.15

pH= 10.85