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?
1 Answer
Explanation:
The first thing to do here is determine the concentrations of the weak acid, which in your case is acetic acid,
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
#["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 = 1.8 * 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
#"pH" = - log(1.8 * 10^(-5)) + log( (0.122 color(red)(cancel(color(black)("M"))))/(0.167color(red)(cancel(color(black)("M")))))#
#"pH" = 4.74 + (-0.14) = color(green)(4.6)#
Now, when a strong acid like hydrochloric acid,
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
#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#
the reaction will look like this
#"CH"_3"COO"_text((aq])^(-) + "H"_3"O"_text((aq])^(+) -> "CH"_3"COOH"_text((aq]) + "H"_2"O"_text((l])#