In: Chemistry
A. 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.
Ka of acetic acid = 1.8 x 10^-5
Acetic acid/sodium acetate buffer
Acetic acid .1M
Sodium acetate .1M
first consider HCl added to pure water
we know that
moles = molarity x volume (L)
so
moles of HCl added = 0.1 x 0.25 x 10-3 = 2.5 x 10-5
now
final volume = 10 + 0.25 = 10.25 ml
now
[HCl ] = moles / volume (L)
[HCl] = 2.5 x 10-5 / 10.25 x 10-3
[HCl ] = 2.44 x 10-3
we know that
HCl is a very strong acid
so
100 % dissociation
HCl --> H+ + Cl-
so
[H+] = [HCl] dissocaited = 2.44 x 10-3
now
the hydronium ion concentration is 2.44 x 10-3
2)
now consider the buffer
the buffer contains an acid , CH3COOh and base Ch3COO-
so
when HCl is added
it reacts with the base CH3C00-
so
the reaction is
H+ + CH3C00- --> CH3COOH
now
initially
moles of CH3C00- = 0.1 x 10 x 10-3 = 1 x 10-3
moles of Ch3COOH = 0.1 x 10 x 10-3 = 1 x 10-3
now
moles of HCl added = 0.25 x 0.1 x 10-3 = 0.025 x 10-3
consider the reaction
H+ + Ch3OO- --> CH3COOH
moles of CH3C00- reacted = moles of H+ added = 0.025 x 10-3
moles ofo CH3C00- remaining = 0.975 x 10-3
moles of CH3COOH formed = moles of H+ added = 0.025 x 10-3
moles of CH3COOH finally = 1.025 x 10-3
now
pH = pKa + log [ CH3C00- / Ch3COOH]
pH = -log 1.8 x 10-5 + log [ 0.975 x 10-3 / 1.025 x 10-3 ]
pH = 4.723
now
-log [H+] = 4.723
[H+] = 1.89 x 10-5