Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 34 mL of 0.16 M C₆H₅OH(aq) with 0.16 M NaOH(aq). The K_a of phenol is 1.0 x 10^-10

^{The answer is 11.45 I'm just not sure how to get it}

Answer 1

find the volume of NaOH used to reach equivalence point

M(C6H5OH)*V(C6H5OH) =M(NaOH)*V(NaOH)

0.16 M *34.0 mL = 0.16M *V(NaOH)

V(NaOH) = 34 mL

we have:

Molarity of C6H5OH = 0.16 M

Volume of C6H5OH = 34 mL

Molarity of NaOH = 0.16 M

Volume of NaOH = 34 mL

mol of C6H5OH = Molarity of C6H5OH * Volume of C6H5OH

mol of C6H5OH = 0.16 M * 34 mL = 5.44 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.16 M * 34 mL = 5.44 mmol

We have:

mol of C6H5OH = 5.44 mmol

mol of NaOH = 5.44 mmol

5.44 mmol of both will react to form C6H5O- and H2O

C6H5O- here is strong base

C6H5O- formed = 5.44 mmol

Volume of Solution = 34 + 34 = 68 mL

Kb of C6H5O- = Kw/Ka = 1*10^-14/1*10^-10 = 1*10^-4

concentration ofC6H5O-,c = 5.44 mmol/68 mL = 0.08M

C6H5O- dissociates as

C6H5O- + H2O -----> C6H5OH + OH-

0.08 0 0

0.08-x x x

Kb = [C6H5OH][OH-]/[C6H5O-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1*10^-4)*8*10^-2) = 2.828*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

1*10^-4 = x^2/(8*10^-2-x)

8*10^-6 - 1*10^-4 *x = x^2

x^2 + 1*10^-4 *x-8*10^-6 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 1*10^-4

c = -8*10^-6

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 3.201*10^-5

putting value of d, solution can be written as:

x = {-1*10^-4 + √(3.201*10^-5)}/2

x = {-1*10^-4 - √(3.201*10^-5)}/2

solutions are :

x = 2.779*10^-3 and x = -2.879*10^-3

since x can't be negative, the possible value of x is

x = 2.779*10^-3

[OH-] = x = 2.779*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.779*10^-3)

= 2.55

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.55

= 11.45

Answer: 11.45