Question

A `6.0*g` mass of acetic acid in `1*L` of water solution is prepared. What mass of sodium acetate is required to give a `pH=4.74`?

Answer 1

Approx. `8*g` of sodium acetate are added.....

Explanation:

....`pH=pK_a+log_10{[[AcO^-]]/[[HOAc]]}`...we want `pH=4.74`, and `pK_a=4.76`....

And thus, clearly, `log_10{[[AcO^-]]/[[HOAc]]}=-0.02`...

And so...`[[AcO^-]]/[[HOAc]]=10^(-0.02)=0.955`...

`"Moles of acetic acid"=(6.0*g)/(60.05*g*mol^-1)=0.0999*mol`...and thus when dissolved in `1*L` we have a concentration of `0.0999*mol*L^-1`.

`[AcO^(-)]=0.955xx0.0999*mol*L^-1=0.0954*mol`

And thus we add a mass of `0.0954*molxx82.03*mol^-1=7.83*g` of sodium acetate.

The molar quantities of sodium acetate and acetic acid are very similar....and thus the `pH` expressed by the buffer is VERY CLOSE to `pK_a""_(HOAc)`.........