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

Explanation / Answer

Moles of HCl added: 0.25/1000 x 0.10 = 2.5x10-5 moles Total Volume = 10 + 0.25 = 10.25 mL or 0.01025 L.

Now for water:

[H3O+] = 2.5x10-5 / 0.01025 = 0.00244 M

pH = -log(0.00244) = 2.61

For the buffer solution:

CH3COONa + HCl --------> CH3COOH + NaCl

moles of CH3COOH = moles of HCl + innitial moles of CH3COOH = 2.5x10-5 + (0.1*0.010) = 0.001025 moles

moles of CH3COONa = 2.5x10-5 - (0.1*0.010) = 0.000975 moles

pH = pKa + log(mol of CH3COONa / mol of CH3COOH)

pH = -log(1.8x10-5) + log (0.000975 / 0.001025)

pH = 4.74 + 0.02 = 4.76

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