# A buffer was prepared by adding 20g of acetic acid CH_3COOH and 20g sodium acetate to enough water to make a 2.0L solution.Determine the pH and write an equation for reaction when a few drops of HCl are added to the buffer?

Feb 9, 2016

$\text{pH} = 4.6$

#### Explanation:

The first thing to do here is determine the concentrations of the weak acid, which in your case is acetic acid, $\text{CH"_3"COOH}$, and of the conjugate base, which in your case is the acetate anion, ${\text{CH"_3"COO}}^{-}$.

Start by using the molar masses of the weak acid and of the salt of the conjugate base, which is of course sodium acetate, to find how many moles of each you're adding to the solution

20 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "0.333 moles CH"_3"COOH"

20color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COONa")/(82.03color(red)(cancel(color(black)("g")))) = "0.244 moles CH"_3"COONa"

Use the volume of the buffer solution to find the concentrations of the two chemical species that are of interest - remember that sodium acetate dissociates in a $1 : 1$ mole ratio with the acetate anion

["CH"_3"COOH"] = "0.333 moles"/"2.0 L" = "0.167 M"

["CH"_3"COO"^(-)] = "0.244 moles"/"2.0 L" = "0.122 M"

Now, the pH of a buffer solution can be calculated using the Henderson - Hasselbalch equation, which looks like this

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

Look up the value of the acid dissociation constant, ${K}_{a}$, for acetic acid. You'll find it listed as

${K}_{a} = 1.8 \cdot {10}^{- 5}$

http://www.bpc.edu/mathscience/chemistry/table_of_monoprotic_acids.html

Plug in your values to solve for the pH of this buffer solution. Notice that since you have a higher concentration of weak acid than of conjugate base, you should expect the pH to be lower than $p {K}_{a}$.

"pH" = - log(1.8 * 10^(-5)) + log( (0.122 color(red)(cancel(color(black)("M"))))/(0.167color(red)(cancel(color(black)("M")))))

$\text{pH} = 4.74 + \left(- 0.14\right) = \textcolor{g r e e n}{4.6}$

Now, when a strong acid like hydrochloric acid, $\text{HCl}$, is added to the buffer, the conjugate base will react with the hydronium ions released by the acid in solution in a neutralization reaction.

This reaction will produce water and acetic acid, which is of course a weak acid. As a result, the pH of the solution will not change significantly by the addition of the strong acid.

Since you have

${\text{HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

the reaction will look like this

${\text{CH"_3"COO"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> "CH"_3"COOH"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$