Question

0.82 g of Sodium Acetate (NaAc) is filled with DI water up to 100 mL. 2.3 mL Acetic Acid is also filled with DI water up to 100 mL. 47.5 mL HAc (0.10 M) and 2.5 mL NaAc (0.10M) are mixed. a) What is the pH of the solution after HAc and NaAc are mixed? b) What is the pH value if 2.0 mL NaOH (0.10M) is added to the solution?

Answer 1

NaAc moles initially = mass / Molar mass of NaAc = ( 0.82g) / ( 82g/mol) = 0.01 mol

NaAc moles added later = M x V ( inL) = 0.1 x 0.0025 = 0.00025

total NaAc moles = 0.01+0.00025 = 0.01025

Acetic acid mass initiallly = volume x density of acid = 2.3ml x 1.05g/ml = 2.415g

acetic acid moles = mass / molar mass of acetic acid = ( 2.415g) / ( 60.05g/mol) = 0.04025 mol

acetic acid moles added later = M x V ( inL) = 0.1 x 0.0475 = 0.00475

total acid moles = 0.04025+0.00475 = 0.045

a) we use Henderson eq to get pH , since we had buffer a weak acid ( acetic acid) and its conjugate base ( sodium acetate)

pH = pka + log ( conjugate base moles) / ( acid moles)

= 4.75 + log ( 0.01025 / 0.045) = 4.11

b) NaOH moles added = M x V ( inL) = 0.1 x 0.002 = 0.0002

we have reacion CH3COOH + NaOH <--> CH3COONa + H2O

thus after adding NaOH acetic acid moles = 0.045-0.0002 = 0.0448

NaAc moles = 0.01025 + 0.0002 = 0.01045 mol

pH = 4.75 + log ( 0.01045 /0.0448)

= 4.12