Question

Please give your pH answers to two decimal places.

A 100.0mL 0.100M weak acid solution is titrated with a 0.100M NaOH solution. If the acid has a Ka of 3.4 x 10^-5, what is the pH of the acid solution...

Before any NaOH is added =

At the equivalence point in the titration =

Answer 1

1)when 0.0 mL of NaOH is added

HA dissociates as:

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

0.1 0 0

0.1-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 ((3.4*10^-5)*0.1) = 1.844*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

3.4*10^-5 = x^2/(0.1-x)

3.4*10^-6 - 3.4*10^-5 *x = x^2

x^2 + 3.4*10^-5 *x-3.4*10^-6 = 0

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

a = 1

b = 3.4*10^-5

c = -3.4*10^-6

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 = 1.36*10^-5

roots are :

x = 1.827*10^-3 and x = -1.861*10^-3

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

x = 1.827*10^-3

use:

pH = -log [H+]

= -log (1.827*10^-3)

= 2.7383

Answer: 2.74

2)

find the volume of NaOH used to reach equivalence point

M(HA)*V(HA) =M(NaOH)*V(NaOH)

0.1 M *100.0 mL = 0.1M *V(NaOH)

V(NaOH) = 100 mL

Given:

M(HA) = 0.1 M

V(HA) = 100 mL

M(NaOH) = 0.1 M

V(NaOH) = 100 mL

mol(HA) = M(HA) * V(HA)

mol(HA) = 0.1 M * 100 mL = 10 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HA) = 10 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react to form A- and H2O

A- here is strong base

A- formed = 10 mmol

Volume of Solution = 100 + 100 = 200 mL

Kb of A- = Kw/Ka = 1*10^-14/3.4*10^-5 = 2.941*10^-10

concentration ofA-,c = 10 mmol/200 mL = 0.05M

A- dissociates as

A- + H2O -----> HA + OH-

0.05 0 0

0.05-x x x

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

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 ((2.941*10^-10)*5*10^-2) = 3.835*10^-6

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

so, x = 3.835*10^-6 M

[OH-] = x = 3.835*10^-6 M

use:

pOH = -log [OH-]

= -log (3.835*10^-6)

= 5.4163

use:

PH = 14 - pOH

= 14 - 5.4163

= 8.5837

Answer: 8.58