Question

Calculate the mass of sodium acetate that must be added to 200.0 ml of a 0.20 M solution of acetic acid to prepare a buffer with a pH value of 4.75? The addition of sodium acetate does not change the overall volume of the solution. (Ka= 1.8 x 10^-5)

I am looking for the steps showing how to solve this problem. Thank you! :)

Answer 1

3.32 g

Explanation:

Ethanoic acid is a weak acid and dissociates:

`sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)`

For which `sf(K_a=([CH_3COO^-][H^+])/([CH_3COOH])=1.8xx10^(-5))`

Please remember that these are equilibrium concentrations .

`sf(pK_a=-logK_a=-log[1.8xx10^(-5)]=4.74)`

`sf(n=cxxv)`

`:.` `sf(n_(CH_3COOH)=0.20xx200.0/1000=0.04)`

This is the initial moles of ethanoic acid. This will be slightly less than the equilibrium moles. Because the value of `sf(K_a)` value is so small we can make the assumption that the equilibrium moles of the species involved is a good approximation to the initial moles.

This is an important assumption which is often not stated. As a guide this assumption is valid for a `sf(pK_a)` range of 4 - 9.

Since `sf(pH=4.75)` then `sf(-log[H^+]=4.75)`

From this `sf([H^+]=1.778xx10^(-5)color(white)(x)"mol/l")`

Rearranging the expression for `sf(K_a)` we get:

`sf([H^(+)]=K_axx([CH_3COOH])/([CH_3COO^(-)]))`

`sf(1.778xx10^(-5)=1.8xx10^(-5)xx0.04/n_(CH_3COO^(-)))`

Note I am using moles instead of concentrations. I can do this because the total volume is common to both acid and co base so just cancels.

This means that the assumption that the question asks you make regarding no volume change is actually irrelevant.

Rearranging:

`sf(n_(CH_3COO^(-))=(1.8xxcancel(10^(-5))xx0.04)/(1.778xxcancel(10^(-5)))=0.4045)`

`sf(m=nxxM_r=0.0405xx82.03=3.32color(white)(x)g)`