Question

0 Mava=mV6 XvP Soma = Moub - Vaan. Determination of the castante Weak Acid D. Concentration of Unknown Acid Trial 3 - Trial Trial Volume of unknown acid Average molarity of NaOH from above mL of NaOH at equivalence point Molarity of unknown acid Average molarity (show calculations) Standard deviation QUESTIONS 1. What are the largest sources of error in this experiment? 2. What is the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH? We assume additive volumes 3. What is the pH of a solution that is 0.75 M in sodium acetate and 0.50 M in acetic acid? (K, for acetic acid is 1.8x10 )

Answer 1

Q2:

Millimole of NaOH = 30.00ml×0.125mol/L = 3.75mmol

Millimole of HCl = 30.00ml× 0.250mol/L = 7.50mmol

Millimole of HCl left after neutralization by NaOH = 7.50 - 3.75 = 3.75 mmol

[HCl] = 3.75mmol/60.00ml = 0.0625mol/L

pH = - log[H⁺] = -log(0.0625) = 1.20412

pH = 1.204 (Answer)

Q3:

A solution of sodium acetate and acetic acid is a buffer solution .

pH = pKa + log([NaOAc]/[AcOH]) = - log(1.8×10^-5) + log(0.75/0.50) = 4.7447 + 0.1761

= 4.9208

pH = 4.92 (Answer)