Question

You are instructed to create 100 mL of a 0.05 M Acetate buffer with a pH of 4.2. You have Glacial Acetic acid (17.4M) and the sodium Acetate NaC2H302, available (Ka=1.8e-5). (Enter all numerical answers to three significant figures.) Copy the results into your lab notebook for use in class. HC2H302(s) + H20(I) = H30+(aq) + C2H302 (aq) What is the molarity needed for the acid component of the buffer? What is the molarity needed for the base component of the buffer? 4. 02 How many moles of acid are needed for the buffer? mol How many moles of base are needed for the buffer? mol How many mL of Glacial Acetic acid are needed for the buffer? 497 5 mL How many grams of base are needed for the buffer? 496

Answer 1

Answer

Molarity needed for the acid component of the buffer = 0.0388M

Molarity needed for the base component of the buffer = 0.00112M

Moles acid needed for the buffer = 0.00388mol

Moles of base needed for the buffer = 0.00112mol

mL of Glacial Acetic acid needed for the buffer = 0.223ml

Grams of base are needed for the buffer = 0.0919g

Explanation

Henderson-Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

4.2 = 4.74 + log( [CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) = -0.54

[CH3COO^-]/[CH3COOH] = 0.2884

[CH3COO^-] = 0.2884[CH3COOH]

[CH3COO^-] + [CH3COOH] = 0.050M

0.2884[CH3COOH] + [CH3COOH] = 0.050M

1.2884[CH3COOH] = 0.050M

[CH3COOH] = 0.0388M

[CH3COO^-] = 0.050M - 0.03881M = 0.0112M

moles of CH3COOH required = (0.03881mol/1000ml)×100ml = 0.00388mol

moles of CH3COO^- required = ( 0.01119mol/1000ml)× 100ml = 0.00112mol

volume of Glavial acetic acid needed = (1000ml/17.4mol)× 0.00388mol = 0.223ml

mass of base required = 0.00112mol × 82.04g/mol = 0.09188g