Question

Calculate the pH at the equivalence point in titrating 0.047 M solutions of each of the following with 0.055 M NaOH (a) nitric acid (HNO3) pH= (b) аcetic acid (HC2H302), ка = 1.8e-05 pH = (c) benzoic acid (HC7H502), Ka 3 6.Зе-05 pH =

Calculate the pH at the equivalence point in titrating 0.047 M solutions of each of the following with 0.055 M NaOH.

(a) nitric acid (HNO₃)

pH =



(b) acetic acid (HC₂H₃O₂), K_a = 1.8e-05

pH =



(c) benzoic acid (HC₇H₅O₂), K_a = 6.3e-05

pH =

Answer 1

1)

Concentration of HNO3 = 0.047M

concentration of NaOH = 0.055M

HNO3 and NaOH are strong acids and strong bases

Hence, PH=7

2)

Concentration of acetic acid= ( HC2H3O2) = 0.047M

Concentration of NaOH = 0.055M

Ka = 1.8 x10^-5

at equivalent point

HC2H3O2 ------------- H+ + C2H3O2-

at equivalent point

Concentration of C2H3O2- = [HC2H3O2][NaOH] / [HC2H2O2]+[NaOH]

[ C2H3O2- ] = 0.047 x 0.055 / 0.047+0.055 = 0.0253M

[ C2H3O2- ] = 0.0253M

C2H3O2- + H2O ------------------- HC2H3O2 + OH-

0.0253        0 0

-x    +x    +x

(0.0253-x)                    +x +x

Ka = 1.8 x10^-5

Ka xKb = Kw    where Kw = ionic product of water = 1.0x10^-14

Kb = Kw / Ka = 1.0x10^-14 / 1.8x10^-5 = 5.55 x10^-10

Kb = 5.55 x10^-10

Kb = [ HC2H3O2] [ OH-] / [C2H3O2-]

5.55 x10^-2 = x*x /( 0.0253 -x )

for solving the equation

x= 3.75 x10^-6

[OH-] = 3.75x10^-6

-log[OH-] = -log( 3.75 x10^-6)

POH = 5.42

PH + POH = 14

PH = 14 - POH

PH = 14 - 5.42

PH = 8.58

c) Benzoic acid = ( HC7H5O2) = 6.3x10^-5

Concentration of acid = 0.047M

Concetration of base = 0.055

Ka = 6.3x10^-5

HC7H5O2 ------------------- H+ + C7H5O2-

at equivalent point

[ C7H5O2- ] = Mbase x Macid / Mbase + Macid

[ C7H5O2- ] = 0.047 x 0.055 / 0.047+0.055

[ C7H5O2- ] = 0.0253M

C7H5O2-   + H2O ------------------------- HC7H5O2 + OH-

0.0253                                             0                 0

-x                                                   +x                +x

0.0253-x                                        +x                  +x

Kb = Kw/ka = 1.0x10^-14 / 6.3x10^-5 = 1.59 x10^-10

Kb = 1.59 x10^-10

Kb = [ HC7H5O2] [ OH-]/[ C7H5O2- ]

1.59x10^-10 = x*x/(0.0253-x)

for solving the equation

x= 2.00 x10^-6

[OH-] = 2.00x10^-6

-log[OH-] = -log(2.00 x10^-6)

POH = 5.70

PH = 14 - 5.70

PH = 8.3