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 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 =

Answer #1

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**

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