One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buff

Question

One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that “one solution is a buffer” is not sufficient. H3O+(aq) + C2H3O2-(aq) HC2H3O2(aq) +H2O Ka of acetic acid = 1.8 x 10-5 Acetic acid/sodium acetate buffer: Acetic acid = 0.1M, Sodium acetate = 0.1M

show work

Explanation / Answer

pKa = -log[1.8*10^-5]

= 4.74

If both are present in equal amount means [A-]=[HA] =1.0

pH = pKa + log[A-/HA]

pH = 4.74+0

pH = 4.74

now add HCl = 2.5*10^-4 L * 0.10 = 2.5*10^-5 moles

1.0 moles sodium acetate + 2.5*10^-5 moles HCl = 1.000025 moles acetic acid.
1.0 moles sodium acetate - 2.5*10^-5 moles = 0.999975 moles sodium acetate.

Molarity of acid = 1.000025 moles / 0.01025 L=97.56 3M

Molarity of acetate= 0.999975 moles / 0.01025 L=97.559 M

pH = pKa + log[A-/HA]

pH = 4.74+ log 97.559/97.563

pH = 4.74

Amount of HCl = now add HCl = 2.5*10^-4 L * 0.10 = 2.5*10^-5 moles

Concentration = amount(mol)/volume(L) = 2.5*10^-5 moles /0.01025 L

= 2.44*10^-3 mol/L or M

Concentration of H+ is also 2.44*10^-3 M as HCl is a strong acid

pH = -log[H+] = -log(2.44*10^-3)

= 2.612

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.