Question 1 Calculate the concentrations of acetic acid and sodium acetate in the

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Question 1

Calculate the concentrations of acetic acid and sodium acetate in the buffer solution you will prepare in this experiment:

Acetic acid concentration is Blank .01 M (2 dec places)

Sodium acetate concentration is Blank .01 M (2 dec places)

The theoretical pH of this buffer solution is (hint: use Henderson-Hasselbach) is 4.74 (2 dec places).

Question 2

Which reaction occurs when you add NaOH to the buffer solution? (Ac = acetate)

OH-   +   H3O+ 2 H2O

Ac- + H3O+ HAc + H2O

HAc   + H3O+   Ac- + H2O

HAc + Ac- Ac- + HAc

Ac- + OH- AcOH

HAc + OH- Ac- + H2O

HAc + H3O+ H2 + H2O + Ac-

Question 3

What is the pH of a solution formed by adding 2 X 10-3 mol of NaOH to 100 ml of the buffer described in question (1)? The calculated pH value is Blank 1 (2 decimal places).

Question 4

Which reaction occurs when you add HCl to the buffer solution? (Ac = acetate) (H30+ = acid)

Ac- + OH- AcOH

HAc   + H3O+   Ac- + H2O

HAc + OH- Ac- + H2O

HAc + H3O+ H2 + H2O + Ac-

OH-   +   H3O+ 2 H2O

HAc + Ac- Ac- + HAc

Ac- + H3O+ HAc + H2O

Question 5

If I put a drop of thymol blue into the above buffer solution, the solution color would be Blank 1. If I put a drop of bromothymol blue in the above buffer solution, the solution color would be Blank 2. If I put a drop of phenolphthalein into the above buffer solution, the solution color would be Blank 3.

a.

OH-   +   H3O+ 2 H2O

b.

Ac- + H3O+ HAc + H2O

c.

HAc   + H3O+   Ac- + H2O

d.

HAc + Ac- Ac- + HAc

e.

Ac- + OH- AcOH

f.

HAc + OH- Ac- + H2O

g.

HAc + H3O+ H2 + H2O + Ac-

Explanation / Answer

pKa of acetic acid/acetate buffer = 4.74

pH = pKa + log [acetate/acetic acid]

If the concentration of acetic acid and acetate is same, pH = pKa. So, the concentration of acetic acid = 0.01M and concentration of sodium acetate is 0.01M

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2. (f) : HAc + OH- Ac- + H2O

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NaOH will react with acetate to convert it to acetate.

Moles of acetic acid intially present = 0.01 moles (If you have a 1L solution of buffer)

moles of acetic acid present after adding NaOH = 0.01-0.002 = 0.008 moles

Moles of acetate present = 0.01 moles + 0.002 moles = 0.012 moles

pH = pKa + log [acetate/acetic acid]

     = 4.74 + log [0.012/0.008] = 4.92

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4. g.

Ac- + H3O+ HAc + H2O

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If I put a drop of thymol blue into the above buffer solution, the solution color would be yellow. If I put a drop of bromothymol blue in the above buffer solution, the solution color would be yellow. If I put a drop of phenolphthalein into the above buffer solution, the solution color would be colorless.

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