If you add 5.00 mL of 0.100 M sodium hydroxide to 50.0 mL of acetate buffer that is 0.100 M in both acetic acid and sodium acetate, what is the pH of the resulting solution? Acetic Acid: Ka = 1.8. x 10-5

see explanation ...

Explanation:

5.00ml (0.100M NaOH) + 0.100M HOAc/NaOAc Bfr

5.00ml (0.100M NaOH) = 0.005 (0.100) mole NaOH = 0.0005 mole NaOH = 0.0005 mole OH⁻ in 55 ml Bfr solution (50ml + 5ml) = > [OH⁻] = (0.0005/0.055) M OH⁻ = 0.0091M OH⁻ ≈ 0.010M OH⁻ added into Bfr solution. The amount of OH⁻ added must be removed by H⁺ in the HOAc equilibrium; that is, H⁺ + OH⁻ → H₂O leaving a void at the H⁺ position in the HOAc equilibrium. HOAc then decomposes to replace the H⁺ removed by the excess OH⁻ giving new H⁺ and OAc⁻ concentrations. Reaction shifts right = > subtract 0.01M from HOAc side of equilibrium and add 0.01M to OAc⁻side of equilibrium and recompute the H⁺ concentration and new pH.

HOAc ⇄ H⁺ + OAc⁻

C (i) 0.10M ~0M * 0.10M = >pH=-log (Ka) = -log (1.85x10⁻⁵) = 4.73

ΔC - 0.01M + x + 0.01M

C (eq) 0.09M x 0.11M = > New HOAc equil. conc.

Ka = [H⁺][OAc⁻]/[HOAc]

=> x (0.11) / (0.09) = 1.85x10⁻⁵

=> x = [H⁺] = 0.09 (1.85x10⁻⁵) / 0.11 = 1.52x10⁻⁵M (after adding NaOH)

=> pH = - log[H⁺] - log (1.52x10⁻⁵) = 4.82 (pH shifts to more basic value b/c of OH⁻ addition)