Calculate the volume of glacial acetic acid (17.6 M) and mass of sodium acetate

Question

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 times 10^-5. (this question is challenging, use the background section and your book for help). 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). A buffer solution is prepared by mixing 20.0 mL of 0.032 M benzoic acid C_6H_5COOH(aq) solution with 25.0 mL of 0.022 M NaC_6H_5COO_(aq). What is the pH of the buffer produced? (K_a of benzoic acid = 6.2 times 10^-5).

Explanation / Answer

Q1.

Vacid for a buffer:

the equation for a buffer:

pH = pKa + log(A-/HA)

pKa = -log(Ka) = -log(1.4*10^-5) = 4.85387

mol of acid = MV = 17.6*V

mol of acetate = mass/MW = x/82.03

V = 102 mL of 0.2 M buffer

total buffer mol

MV = 102*0.2 = 20.4 mmol = 20.4*10^-3 = 0.0204

pH = 4.85 so:

pH = pKa + log(A-/HA)

4.85 = 4.85387 + log(A-/HA)

A-/HA = 10^(4.85-4.85387) = 0.99112

A- = HA*0.99112

now..

A- + HA = 0.0204

HA*0.99112 + HA = 0.0204

1.99112*HA = 0.0204

HA = 0.0204/1.99112 = 0.0102454 mol

A- = HA*0.99112 = 0.0102454 *0.99112 = 0.010154

mass of salt = mol*MW = 0.010154*82.03 = 0.83293262 g of Sodium Acetate

for acid:

M = mol/V

V = mol/M = 0.0102454 /17.6 = 0.000582125 liters = 0.582 mL

This must be diluted to V = 102 mL of ionized water

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