Question

please explain steps

1. Calculate the volume of glacial acetic acid (17.6 M) and mass of sodium acetate (mm-82.03) required to make 102 ml of 0.20 M buffer at pH 4.85. The Ka of acetic acid is 1.4 x10. (this question is challenging, use the background section and your book for help). 2. Consult the table of Ka values in the back of your textbook and suggest acid/conjugate base system (mono-protic acid only) that would yield a strong buffer capacity at pH of 8.60 (the buffer must be strong compared to others of the same molarity). Explain your selection (simply and succinctly). 3. A buffer solution is prepared by mixing 20.0 mL of 0.032 M benzoic acid C H5COOH solution with 25.0 mL of 0.022 M NaC&H COO(ag- What is the pH of the buffer produced? (Kg of benzoic acid =6.2 x10)

Answer 1

1-

A buffer is a solution where both a weak acid (here acetic acid) and the salt of its conjugate base (here sodium acetate) is present. Now pH of a buffer is calculated by Henderson Hasselbalch equation

pH = pKa + log [salt]/[acid]

Where pKa = -log Ka and Ka is the dissociation constant of acid at equilibrium.

Now given the total concentration of buffer = [salt] + [acid] = 0.20 M

Total volume of the buffer = 102 ml

pH of the buffer = 4.85

pKa = -log Ka = -log (1.4 * 10^-5) =  5 - log (1.4) = 5 - 0.146 = 4.854‬

Now lets consider [salt] = xM

Then [acid] = 0.20 M - xM

Now putting these values in the equation-

pH = pKa + log [salt]/[acid]

4.85 = 4.854‬ + log [xM]/[0.20 M - xM]

log [xM]/[0.20 M - xM] = 4.85 - 4.854‬ = -0.004

[xM]/[0.20 M - xM] = 10^-0.004 = 0.99

[xM] = 0.99 * [0.20 M - xM]

x = 0.198 - 0.99x

x + 0.99x = 0.198

1.99 x = 0.198

x = 0.198 / 1.99

x = 0.099 M

Thus [salt] = xM = 0.099 M

[acid] = 0.20 M - xM = 0.20 M - 0.099 M = 0.101 M

c-

Now the total volume of solution (V2)= 102 mL

Final concentration of acetic acid (M2) = 0.101 M

Initial concentration of acetic acid (M1) = 17.6 M

Thus initial volume of acetic acid taken = V1 = M2V2/M1

= 0.101 M * 102 mL / 17.6 M

= 0.585 mL

d-

Similarly concentration of sodium acetate = 0.099 M

Volume of total solution = 102 mL

Thus moles of sodium acetate taken = concentration * volume

= 0.099 M * 102 mL

= 0.099 mol/1000 ml * 102 mL

= 0.01 mols

Again given molar mass of sodium acetate = 82.03 g/mol

Thus mass of sodium acetate taken = moles * molar mass

= 0.01 mols * 82.03 g/mol

= 0.8203 g