Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.90-M HCl to 350. mL of each of the following solutions.

Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

a) water

pH before mixing =
pH after mixing=
pH change =

b) 0.118 M C₂H₃O₂^1-

pH before mixing =
pH after mixing=
pH change =

c) 0.118 M HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.118 M in each C₂H₃O₂^1- and HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

Answer 1

1)pH of water (before mixing) = 7.00

After molarity is calculated as M1V1 = M2V2

(350+10) x M1 = 10 x 3.90

M1 = 0.1083 M

and pH = -log 0.1083 =0.9652

thus change in pH = 7.00-0.9652 = 6.0347

2) before mix , the solutionis of salt of weak acid and strong base , whose pH is

pH = 1/2 [pkw + pKa + log C]

= 1/2 [14 + 4.74 +log 0.118]

=8.9059

After mixing

A- + HCl -----------> HA + Cl-

350x0.118 0 0 0 initial

- 10x 3.9 - - change

2.3 0 39 - after mmoles

It is a buffer whose pH is by Hendersen equation

pH = pKa + log [Conjugate base]/[acid]

= 4.74 + log 2.3/39

=3.510

thus chang ein pH = 8.9059 - 3.510

= 5.3953

3) the solution is of weak acid whose pH = 1/2 [pKa - logC]

pH = 1/2[ 4.74 - log 0.118]

= 2.834

After additing 10 mL of 3.9 Hcl, a strong acid, due to common ion effect the dissociation of acetic acid isvery much decrease, hence all [H+] from Hcl .thus

pH = - log [H+] from Hcl

= 0.9652

thus chang ein pH = 2.834 - 0.9652

= 1.869

4) A buffer wit [HA] = [A-]

Thus its pH = pKa = 4.74

after adding Hcl

  

A- + HCl -----------> HA + Cl-

350x0.118 0 350x0.118 0 initial

- 10x 3.9 - - change

2.3 0 80.3 - after m moles

pH = 4.74 + log 2.3/80.3

= 3.19

Thus change in pH = 1.543