Question

Phenol (C6H5OH) has a Ka = 1.05 x 10-10. If 100.0 mL of a 0.5000 M aqueous phenol solution is mixed with 100.0 mL of 0.5000 M aqueous sodium hydroxide, the resulting solution will have a pH

CORRECT ANS: >7

Answer 1

we have:

Molarity of C6H5OH = 0.5 M

Volume of C6H5OH = 100 mL

Molarity of NaOH = 0.5 M

Volume of NaOH = 100 mL

mol of C6H5OH = Molarity of C6H5OH * Volume of C6H5OH

mol of C6H5OH = 0.5 M * 100 mL = 50 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.5 M * 100 mL = 50 mmol

We have:

mol of C6H5OH = 50 mmol

mol of NaOH = 50 mmol

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

C6H5O- here is strong base

C6H5O- formed = 50 mmol

Volume of Solution = 100 + 100 = 200 mL

Kb of C6H5O- = Kw/Ka = 1*10^-14/1.05*10^-10 = 9.524*10^-5

concentration ofC6H5O-,c = 50 mmol/200 mL = 0.25M

C6H5O- dissociates as

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

0.25 0 0

0.25-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 ((9.524*10^-5)*0.25) = 4.88*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)

9.524*10^-5 = x^2/(0.25-x)

2.381*10^-5 - 9.524*10^-5 *x = x^2

x^2 + 9.524*10^-5 *x-2.381*10^-5 = 0

Let's solve this quadratic equation

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

a = 1

b = 9.524*10^-5

c = -2.381*10^-5

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 = 9.525*10^-5

putting value of d, solution can be written as:

x = {-9.524*10^-5 + √(9.525*10^-5)}/2

x = {-9.524*10^-5 - √(9.525*10^-5)}/2

solutions are :

x = 4.832*10^-3 and x = -4.927*10^-3

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

x = 4.832*10^-3

[OH-] = x = 4.832*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (4.832*10^-3)

= 2.4

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.4

= 11.7

Answer: 11.7