pH = pKa + log (base)/(acid)
Let x = mL base
then 100-x = mL acid
millimols base = mL x M
millimols acid = mL x M
Substitute into the HH equation and solve for x = mlL base and 100-x = mL acid. That's it.
You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate.
How much of each solution should be mixed to prepare this buffer?
mL of benzoic acid ?
mL of sodium benzoate?
1 answer
2 answers