Question

Titration of 50.0mL of 0.100M HX (Ka=1.5x10^-5) with 0.100M NaOH. Calculate the pH of

1) initial acid solution

2) buffer formed at the addition of 12.5mL NaOH

3) buffer formed at the addition of 25.0mL NaOH

4) buffer formed at the addition of 37.5mL NaOH

5) solution obtained at the endpoint

Answer 1

1)when 0.0 mL of NaOH is added

HX dissociates as:

HX -----> H+ + X-

0.1 0 0

0.1-x x x

Ka = [H+][X-]/[HX]

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 ((1.5*10^-5)*0.1) = 1.225*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)

1.5*10^-5 = x^2/(0.1-x)

1.5*10^-6 - 1.5*10^-5 *x = x^2

x^2 + 1.5*10^-5 *x-1.5*10^-6 = 0

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

a = 1

b = 1.5*10^-5

c = -1.5*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 = 6*10^-6

roots are :

x = 1.217*10^-3 and x = -1.232*10^-3

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

x = 1.217*10^-3

use:

pH = -log [H+]

= -log (1.217*10^-3)

= 2.9146

Answer: 2.91

2)when 12.5 mL of NaOH is added

Given:

M(HX) = 0.1 M

V(HX) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 12.5 mL

mol(HX) = M(HX) * V(HX)

mol(HX) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 12.5 mL = 1.25 mmol

We have:

mol(HX) = 5 mmol

mol(NaOH) = 1.25 mmol

1.25 mmol of both will react

excess HX remaining = 3.75 mmol

Volume of Solution = 50 + 12.5 = 62.5 mL

[HX] = 3.75 mmol/62.5 mL = 0.06M

[X-] = 1.25/62.5 = 0.02M

They form acidic buffer

acid is HX

conjugate base is X-

Ka = 1.5*10^-5

pKa = - log (Ka)

= - log(1.5*10^-5)

= 4.824

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.824+ log {2*10^-2/6*10^-2}

= 4.347

Answer: 4.35

3)when 25.0 mL of NaOH is added

Given:

M(HX) = 0.1 M

V(HX) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(HX) = M(HX) * V(HX)

mol(HX) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HX) = 5 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess HX remaining = 2.5 mmol

Volume of Solution = 50 + 25 = 75 mL

[HX] = 2.5 mmol/75 mL = 0.0333M

[X-] = 2.5/75 = 0.0333M

They form acidic buffer

acid is HX

conjugate base is X-

Ka = 1.5*10^-5

pKa = - log (Ka)

= - log(1.5*10^-5)

= 4.824

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.824+ log {3.333*10^-2/3.333*10^-2}

= 4.824

Answer: 4.82

4)when 37.5 mL of NaOH is added

Given:

M(HX) = 0.1 M

V(HX) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 37.5 mL

mol(HX) = M(HX) * V(HX)

mol(HX) = 0.1 M * 50 mL = 5 mmol

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

mol(NaOH) = 0.1 M * 37.5 mL = 3.75 mmol

We have:

mol(HX) = 5 mmol

mol(NaOH) = 3.75 mmol

3.75 mmol of both will react

excess HX remaining = 1.25 mmol

Volume of Solution = 50 + 37.5 = 87.5 mL

[HX] = 1.25 mmol/87.5 mL = 0.0143M

[X-] = 3.75/87.5 = 0.0429M

They form acidic buffer

acid is HX

conjugate base is X-

Ka = 1.5*10^-5

pKa = - log (Ka)

= - log(1.5*10^-5)

= 4.824

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.824+ log {4.286*10^-2/1.429*10^-2}

= 5.301

Answer: 5.30

Only 4 parts at a time please