Question

# Calculate the pH of each of the solutions and the change in pH to 0.01 pH...

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 2.68-M HCl to 630. mL of each of the following solutions.

a) water

pH before mixing =

pH after mixing=

pH change =

b) 0.174 M C2H3O21-

pH before mixing =

pH after mixing=

pH change =

c) 0.174 M HC2H3O2

pH before mixing =

pH after mixing=

pH change =

d) a buffer solution that is 0.174 M in each C2H3O21- and HC2H3O2

pH before mixing =

pH after mixing=

pH change =

Moles of HCl = 0.01 L x 2.68 M = 0.0268 Moles

Volume = 630 ml + 10 ml = 640 ml

a. We know, pH = - log [H+]

Initial pH = - log (1x10-7) = 7

Final pH = - log ( 0.0268 x 1000 / 640) = - log (4.19 x 10-2) = 1.38

b. Since it's a base,

Initial pH =14 + log (0.174) = 13.24

After adding 0.0268 Moles of HCl,

(0.63 x 0.174) - (0.0268) = 0.0828

Final pH = 14 + log ( 0.0828 x 1000 / 640) = 14 + log (0.129) = 13.11

c. It's a acid,

Initial pH = - log (0.174) = 0.76

After adding 0.0268 Moles of HCl,

(0.63 x 0.174) + (0.0268) = 0.1364

Final pH = - log ( 0.1364 x 1000 / 640) = 0.67

d.

For buffer solution, pH = pKa + log ([Salt] / [Acid])

pKa = - log Ka = - log (1.75 x 10-5) = 4.76

Initial pH = 4.76 + log (0.174/ 0.174) = 4.76

After adding 0.0268 Moles of HCl,

[CH3COOH] = (0.63 x 0.174) + 0.0268 = 0.1364 M

[CH3COO-] = (0.63 x 0.174) - 0.0268 = 0.0828 M

Final pH = 4.76 + log (0.0828 / 0.1364) = 4.54