To make the pH of a 1.3-liter common ion solution of 5.6 M acetic acid and 6.3 M sodium acetate equal to 5, approximately 2 moles of LiOH are needed.
To find the amount of LiOH needed to make the pH of a 1.3-liter common ion solution of 5.6 M acetic acid and 6.3 M sodium acetate equal to 5, we can use the common ion effect. The common ion in this case is the acetate anion (CH3COO−).
Calculate the initial concentration of the acetate anion:
Initial concentration of acetic acid (HA) = 5.6 M
Initial concentration of sodium acetate (NaA) = 6.3 M
Initial concentration of acetate anion (A−) = 6.3 M (strong electrolyte)
Calculate the initial concentration of hydrogen ions (H+):
Initial concentration of hydrogen ions = 5.6 M (from acetic acid) + 6.3 M (from sodium acetate) = 11.9 M
Use the common ion effect equation to find the concentration of hydrogen ions at equilibrium:
[H+]eq = [H+]i * ([A−]i / [A−]eq)
[H+]eq = 11.9 M * (6.3 M / 6.3 M + 5.6 M)
[H+]eq = 11.9 M * (6.3 M / 12.9 M)
[H+]eq ≈ 4.74 M
Calculate the concentration of acetate anion at equilibrium:
[A−]eq = [A−]i * ([H+]eq / [H+]i)
[A−]eq = 6.3 M * (4.74 M / 11.9 M)
[A−]eq ≈ 3.44 M
Use the formula for the pH:
pH = -log [H+]
pH = -log 4.74 M
pH ≈ 5
Calculate the amount of LiOH needed to neutralize the excess hydrogen ions:
2 LiOH + 4.74 M H+ → 2 Li+ + 4.74 M OH−
2 moles of LiOH needed.