Question

A chemist weighs 0.364 g of 1,2,3 trihydroxy benzene and dissolves it in a 100 mL volumetric flask. Calculate the pH. Ka 1.15 x10-9

mw = 123 g/mol

Answer 1

mass, m = 0.364 g

use:

number of mol,

n = mass/molar mass

=(0.364 g)/(1.23*10^2 g/mol)

= 2.959*10^-3 mol

volume , V = 1*10^2 mL

= 0.1 L

use:

Molarity,

M = number of mol / volume in L

= 2.959*10^-3/0.1

= 2.959*10^-2 M

HA dissociates as:

HA -----> H+ + A-

2.959*10^-2 0 0

2.959*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.15*10^-9)*2.959*10^-2) = 5.833*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.833*10^-6 M

So, [H+] = x = 5.833*10^-6 M

use:

pH = -log [H+]

= -log (5.833*10^-6)

= 5.2341

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)/2.9*10^-8

Kb = 3.448*10^-7

ClO- dissociates as

ClO- + H2O -----> HClO + OH-

0.098 0 0

0.098-x x x

Kb = [HClO][OH-]/[ClO-]

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 ((3.448*10^-7)*9.8*10^-2) = 1.838*10^-4

since c is much greater than x, our assumption is correct

so, x = 1.838*10^-4 M

use:

pOH = -log [OH-]

= -log (1.838*10^-4)

= 3.7356

use:

PH = 14 - pOH

= 14 - 3.7356

= 10.2644

Answer: 10.26