Question

Calculate the mass of sodium acetate (NaC2H3O2) that must be added to make 100 mL of an acetic acid/acetate buffer at pH = 5.0, given that you will use 5.0 mL of 0.50 M acetic acid.

1. Calculate the concentration of acetic acid after 5.0 mL of 0.50 M acetic acid has been diluted to 100.0 mL.

2. Using the Henderson-Hasselbalch equation, calculate the concentration of NaC2H3O2 need in the buffer.

3. Calculate the mass of NaC2H3O2 needed to make 100.0 mL of solution of the concentration you determined in part 2.

4. If you needed a buffer of pH 5.5, would you require more or fewer grams of NaC2H3O2? Why?

PLEASE HELP ME

Answer 1

Reaction taken place:

CH₃COOH + CH₃COONa ? Buffer

pH of buffer = 5

5 = -log[H⁺ion in buffer]

H⁺ion in buffer = Anti-log (-5)

= 10^-5

1).

Initial concentration of acetic acid (M₁) = 0.5 M

Final concentration of acetic acid (M2) = M2

Intial volume of acetic acid (V₁) = 5 ml

Final volmue of acetic acid (V₂) = 100 ml

M1V1 = M2V2

(0.5)*(5) = M2*100

M2 = (0.5)*(5)/100

= 2.5*10^-2 M

2).

pH = pKa + log [product]/[reactant]

pH = 5

pKa = 4.5

pH = pKa + log [Buffer concentration]/[Acid][CH3COONa]

5 = 4.5 + log[[10^-5]/[2.5*10^-2][CH₃COONa]]

log[[10^-5]/[2.5*10^-2][CH₃COONa]]= 5-4.5 = 0.5

[10^-5]/[2.5*10^-2][CH₃COONa]] = Anti-log (0.5) = 3.1622

10^-3/(2.5*3.1622) = [CH₃COONa]

= 1.2*10^-4 grams

3). [CH₃COONa] = (1.2*10^-4) per 100ml

= 1.2*10^-4/100

= 1.2*10^-2

= 0.012 grams of CH₃COONa to be added per 100 ml.

4). From the relation it is to be noted that concentration of CH₃COONa is inversly proportional to pH value, thus inorder to get more pH value, it needs fewer quantity of CH₃COONa.