Question

Calculate the pH at the equivalence point in titrating 0.056 M solutions of each of the following with 0.013 M NaOH.

(a) nitric acid (HNO₃)

pH =

(b) arsenous acid (H₃AsO₃), K_a = 5.1e-10

pH =

(c) ascorbic acid (HC₆H₇O₆), K_a = 8e-05

pH =

Answer 1

1) Concentration of HNO3 = Macid = 0.056M

Concentration of NaOH = Mbase = 0.013M

HNO3 is a strong acid and NaOH is a strong base.

PH =7

b)

Arsenous acid ( H3AsO3)

Ka= 5.1x10^-10

Concentration of acid = Macid = 0.056M

Concentration of base = Mbase = 0.013M

H3AsO3 ---------------------- H+ + H2AsO3-

at equivalent point

Concentration of H2AsO3- = Macid x Mbase / Macid+Mbase

[H2AsO3-] = 0.056 x 0.013 / 0.056+0.013 = 0.0105M

[H2AsO3-] = 0.0105M

                               H2AsO3- + H2O -------------------------- H3AsO3 + OH-

Initial                          0.0105                                               0               0

change                         -x                                                    +x            +x

equilibrium                0.0105-x                                             +x             +x

Ka = 5.1 x10^-10

Ka x Kb = Kw               where Kw = ioinic product of water = 1.0x10^-14

Kb = Kw/Ka = 1.0x10^-14 / 5.1x10^-10 = 1.96 x10^-5

Kb = 1.96x10^-5

Kb = [H3AsO3][OH-]/[H2AsO3-]

1.96 x10^-5 = x*x/( 0.0105-x)

for solving the equation

x= 0.000444

[OH-] = 0.000444M

-log[OH-]= -log(0.000444)

POH = 3.35

PH+POH= 14

PH = 14-POH

PH = 14 - 3.35

PH = 10.65

C)

ascorbic acid ( HC6H7O6)

Ka= 8x10^-5

Concentration of acid = Macid = 0.056M

Concentration of base = Mbase = 0.013M

HC6H7O6-------------------- H+ + C6H7O6-

at equivalent point

concentration of C6H7O6- = Macid x Mbase/Macid + Mbase

[ C6H7O6-] = 0.056 x 0.013 / 0.056+0.013 = 0.0105M

[ C6H7O6-] = 0.0105M

                           C6H7O6- + H2O ---------------------- HC6H7O6 + OH-

inititial                   0.0105                                               0               0

change                  -x                                                    +x            +x

equilibrium          0.0105-x                                             +x             +x

Kb = Kw/Ka = 1.0x10^-14 / 8x10^-5 = 1.25x10^-10

Kb = 1.25 x10^-10

Kb = [ HC6H7O6 ] [ OH-] / [ C6H7O6- ]

1.25x10^-10 = x*x/( 0.0105-x)

for solving the equation

x= 1.145 x10^-6

[OH-] = 1.145x10^-6

-log[OH-]= -log(1.145x10^-6)

POH = 5.94

PH = 14 - 5.94

PH = 8.06