Question #2627e

1 Answer
Oct 29, 2017

Answer:

#"180 g CH"_3"COONa"#

Explanation:

You know that you're dealing with a weak acid-conjugate buffer, so you can say that its #"pH"# can be calculated using this version of the Henderson - Hasselbalch equation.

#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#

Now, you know that the #"pH"# of the buffer is higher than the #"p"K_a# of the weak acid, so right from the start, you can say that

#["CH"_3"COOH"] < ["CH"_3"COO"^(-)]#

This basically means that the buffer contains more conjugate base than weak acid, i.e. more acetate anions--delivered to the solution by the sodium acetate--than acetic acid.

Use the #"pH"# and the #"p"K_a# of the weak acid to find the ratio that exists between the concentration of the acetate anions and the acetic acid in this buffer.

#6 = 4.75 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

This gets you

#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 1.25#

You can rewrite this as

#10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^1.25#

which is equivalent to

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 17.78#

You can thus say that in this solution, you have

#["CH"_3"COO"^(-)] = 17.78 * ["CH"_3"COOH"]#

Now, I assume that you have

#["CH"_3"COOH"] = "0.25 M"#

This means that the concentration of the acetate anions would be

#["CH"_3"COO"^(-)] = 17.78 * "0.25 M"#

#["CH"_3"COO"^(-)] = "4.445 M"#

Next, use the volume of the solution to find the number of moles of acetate anions.

#500 color(red)(cancel(color(black)("mL solution"))) * ("4.445 moles CH"_3"COO"^(-))/(10^3color(red)(cancel(color(black)("mL solution")))) = "2.223 moles CH"_3"COO"^(-)#

As you know, when dissolved in water, sodium acetate dissociates in a #1:1# mole ratio to produce sodium cations and acetate anions, so you can say that in order to have #2.223# moles of acetate anions in the buffer, you need to dissolve #2.223# moles of sodium acetate to make this solution.

Finally, use the molar mass of sodium acetate to find the number of grams of salt.

#2.223 color(red)(cancel(color(black)("moles CH"_3"COONa"))) * "82 g"/(1color(red)(cancel(color(black)("mole NaCH"_3"COONa")))) = color(darkgreen)(ul(color(black)("180 g")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the buffer.