Question

Exactly 25.0 ml of 0.060 M Benzoic acid was placed in a clean dry Erlenmeyer flask and titrated with 0.075 M NaOH solution. Determine the pH of the solution in the flask after each of the following amounts of the NaOH solution is added if the Ka of the benzoic acid is 6.3 x 10^-5

f.) 20.00 ml NaOH

g.) 20.10 ml NaOH

h.) 25.00 ml NaOH

i.) 30.00 ml NaOH

Answer 1

mmol of acid = MV = 0.06*25 = 1.5

Vbase required =  0.06*25/0.075 = 20

then

a)

this is equvialence point

expect hydrolysis

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

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

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

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

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(6.3*10^-5) = 1.587*10^-10

M = mmol/(VtotalI) = 0.06*25 /(25+20) = 0.0333

1.587*10^-10 = x*x/(0.0333-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =2.298*10^-6

[OH-]  =2.298*10^-6

pOH = -log(OH-) = -log(2.298*10^-6 = 5.638

pH = 14-5.638= 8.362

pH = 8.5328

g)

this is excess acid

mmol of H left = (20.10*0.075 - 25*0.06) = 0.0075

Vtotal = 25+20.1 = 45.1

[H+] = 0.0075/ 45.1 = 0.000166

pH = -log(0.000166) = 3.779

h)

this is excess acid

mmol of H left = (25*0.075 - 25*0.06) = 0.375

Vtotal = 25+25=50

[H+] = 0.375/ 50 = 0.0075

pH = -log(0.0075) = 2.1249

i)

this is excess acid

mmol of H left = (30*0.075 - 25*0.06) = 0.75

Vtotal = 25+30=55

[H+] = 0.75/ 55 = 0.0136

pH = -log(0.0136) = 1.8664