Question

Question #2627e

Answer 1

`"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.