# Question 54a99

Feb 4, 2015

Start by determining the concentrations of the two species in the buffer. For acetic acid, you have

C = n/V => n_("acetic") = C * V = "0.2 M" * 100*10^(-3)"L" = 20 * 10^(-3)"moles"

This means that the concentration of acetic acid in the buffer will be

C_("acetic") = n_("acetic")/V_("buffer") = (20*10^(-3)"moles")/((100 + 100)*10^(_3)"L") = "0.1 M"

For sodium acetate, the number of moles is

n_("acetate") = C * V = "0.15 M" * 100 * 10^(-3)"L" = 15 * 10^(-3)"moles"

As a result ,the concentration of sodium acetate in the buffer will be

C_("acetate") = n_("acetate")/V_("buffer") = (15*10^(-3)"moles")/((100 + 100) * 10^(-3)"L") = "0.075 M"

Since you're dealing with a buffer, use the Henderson-Hasselbalch equation to determine the pH of the solution

$p {H}_{\text{solution}} = p K a + \log \left(\frac{\left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}\right)$

$p {H}_{\text{solution") = 4.75 + log("0.075 M"/"0.1M}} = 4.75 - 0.125 = 4.63$

Now you add the $\text{HCl}$ solution. The strong $\text{HCl}$ acid will react with the wak sodium acetate base to produce weak acetic acid

$H C {l}_{\left(a q\right)} + C {H}_{3} C O O N {a}_{\left(a q\right)} \to N a C {l}_{\left(a q\right)} + C {H}_{3} C O O {H}_{\left(a q\right)}$

(I won't go into the net ionic equation)

The moles of hydrochloric acid added to the solution will be

n_("HCl") = C * V = "0.2 M" * 20 * 10^(-3)"L" = 4 * 10^(-3)"moles"

This means that the concentration of $\text{HCl}$ in the buffer will be

C_("HCl") = n_("HCl")/V_("buffer") = (4 * 10^(-3)"moles")/((100 + 100 + 20) * 10^(-3)"L") = "0.018 M"

The new concetrations of acetic acid and sodium acetate will be

C_("acetate") = (15 * 10^(-3)"moles")/(220 * 10^(-3)"L") = "0.068 M", and

C_("acetic") = (20 * 10^(-3)"moles")/(220 * 10^(-3)"L") = "0.091 M"

Now, all the hydrochloric acid will be consumed by the above reaction; this means that the concentration of sodium acetate will decrease by how much $\text{HCl}$ was consumed, and the concentration of acetic acid will increase by the same amount.

Therefore,

C_("acetic-final") = C_("acetic") + C_("HCl") = "0.091 M" + "0.018 M" = "0.109 M", and

C_("acetate-final") = C_("acetate") - C_("HCl") = "0.068 M" - "0.018 M" = "0.05 M"

This means that the solution's pH will now be

pH_("sol") = 4.75 + log(("0.05 M")/("0.109 M")) = 4.75 - 0.338 = 4.41#

Notice how little the pH dropped despite the addition of a strong acid.