Question #2627e

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Question #2627e

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Answer 1 · 2017-10-29T00:18:40

${180 g CH}_{3} COONa$ $"180 g CH"_3"COONa"$

Explanation:

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

$pH = p K_{a} + log (\frac{[conjugate base]}{[weak acid]})$ $"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))$

Now, you know that the $pH$ $"pH"$ of the buffer is higher than the $p K_{a}$ $"p"K_a$ of the weak acid, so right from the start, you can say that

$[{CH}_{3} COOH] < [{CH}_{3} {COO}^{-}]$ $["CH"_3"COOH"] < ["CH"_3"COO"^(-)]$

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 $pH$ $"pH"$ and the $p K_{a}$ $"p"K_a$ of the weak acid to find the ratio that exists between the concentration of the acetate anions and the acetic acid in this buffer.

$6 = 4.75 + log (\frac{[{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]})$ $6 = 4.75 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))$

This gets you

$log (\frac{[{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]}) = 1.25$ $log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 1.25$

You can rewrite this as

$10^{log (\frac{[{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]})} = 10^{1.25}$ $10^log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^1.25$

which is equivalent to

$\frac{[{CH}_{3} {COO}^{-}]}{[{CH}_{3} COOH]} = 17.78$ $(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 17.78$

You can thus say that in this solution, you have

$[{CH}_{3} {COO}^{-}] = 17.78 \cdot [{CH}_{3} COOH]$ $["CH"_3"COO"^(-)] = 17.78 * ["CH"_3"COOH"]$

Now, I assume that you have

$[{CH}_{3} COOH] = 0.25 M$ $["CH"_3"COOH"] = "0.25 M"$

This means that the concentration of the acetate anions would be

$[{CH}_{3} {COO}^{-}] = 17.78 \cdot 0.25 M$ $["CH"_3"COO"^(-)] = 17.78 * "0.25 M"$

$[{CH}_{3} {COO}^{-}] = 4.445 M$ $["CH"_3"COO"^(-)] = "4.445 M"$

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.

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Question #2627e

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