posted by Jematormal91 on .
You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.240 M sodium benzoate.
How much of each solution should be mixed to prepare this buffer?
pH = pKa + log(b/a)
Substitute and I get
b/a= 0.631 so b=0.631*a or
1.585*a = b
Let x = mL of the base(b) and 100-x = mL acid(a)
Then 1.585*mols b = mols a.
1.585(0.240x) = (100-x)*0.1
Solve for x = mL base and 100-x = mL acid.
I get approximately 20 mL base and approximately 80 mL acid but you need to go through and get a better set of numbers. I ALWAYS check it.You should, too.
mmols base = 20 x 0.240 =about 4.8
mmols acid = 80 x 0.1 = about 8.0
pH = 4.2 + log(4.8/8.0) = 4.2 + (-.22) = about 3.98. I think if you put in the correct numbers it will work ok to give 4.00.
31 mL benzoate and 69 mL benzoic acid