Question

The substance phenol (CH3OH) is a weak acid (K4 = 1.0x10-1'). What is the pH of a 0.179 M aqueous solution of sodium phenoxide, NaC,H,O? pH = This solution i acidic basic neutral Submit Ans dy Entire Group 9 more group attempts remaining

Answer 1

1)

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1*10^-10

Kb = 1*10^-4

C6H5O- dissociates as

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

0.179 0 0

0.179-x x x

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

Kb = x*x/(c-x)

Assuming 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)*0.179) = 4.231*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/(0.179-x)

1.79*10^-5 - 1*10^-4 *x = x^2

x^2 + 1*10^-4 *x-1.79*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1*10^-4

c = -1.79*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 7.161*10^-5

roots are :

x = 4.181*10^-3 and x = -4.281*10^-3

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

x = 4.181*10^-3

use:

pOH = -log [OH-]

= -log (4.181*10^-3)

= 2.3787

use:

PH = 14 - pOH

= 14 - 2.3787

= 11.62

Since pH is more than 7, this is basic

Answer: 11.62, basic

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