Question

You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.180 M sodium benzoate

How much of each solution should be mixed to prepare this buffer?

Answer 1

Let x = mL benzoate
Then 100-x = mL acid
--------------------
millimols base = x*0.180
mmoles acid = (100-x)*0.1
4.00 = 4.20 + log(0.180x)/[(100-x)*0.1]
Solve for x and 100-x
I get 26 something for base and about 74 for acid. You should do it more accurately.

Answer 2

I need your helping for assigment of account

Answer 3

Ineed helping full for account