Question

How many grams of sodium acetate, NaC2H3O2, would have to be added to 1.00 L of 0.15 M acetic acid, HC2H3O2, to make a buffer with a pH of 5.00?

Answer 1

Approx. `20*g` of sodium acetate are required....

Explanation:

We use the buffer equation...

A buffer is prepared when a weak base, and its conjugate acid are mixed together in appreciable quantities to give a solution that resists gross changes in `pH`...and for such a system...

`pH=pK_a+log_10{[[A^-]]/[[HA]]}`

For acetic acid `pK_a=4.76`. And so we fill in the dots...

`5.00=4.76+log_10{[[AcO^-]]/[[HOAc]]}`

So here `log_10{[[AcO^-]]/[[HOAc]]}=+0.24`

`{[[AcO^-]]/[[HOAc]]}=10^(0.24)=1.74`

Now we have been given...`[HOAc]=0.15*mol*L^-1`...

Thus...`[AcO^-]=0.15*mol*L^-1xx10^(0.24)=1.74xx0.15*mol*L^-1`

A mass of .....................`1.00*Lxx0.261*mol*L^-1xx82.03*g*mol^-1=21.39*g`.

Let us check this out....

`pH=4.76+log_10{[[0.261*mol*L^-1]]/[[0.15*mol*L^-1]]}=5.00`

As required....