Create a buffer using CH3COOH and NaCH3COO that has a pH of exactly 3.75.?
a 50 mL sample of your buffered solution will have to be able to withstand the addition of 25.0 mL of 0.100 M NaOH solution.
the buffered solution will break after the addition of no more than 35.0 mL of the 0.10 M NaOH.
Can someone give me an overview of how to approach this? Thanks
- davidLv 76 months agoBest answer
pH 3.75 >> [H+] = 10^-3.75 = 1.78x10^-4 M
NaCH3COO --> Na+ + CH3COO- << common ion effect
CH3COOH <--> H+ + CH3COO-
Ka = [H+][CH3COO-] / [CH3COOH]
1.8x10^-5 = (1.78x10^-4)[CH3COO-] / [CH3COOH]
[CH3COO-] / [CH3COOH] = (1.8x10^-5) / (1.78x10^-4)
[CH3COO-] / [CH3COOH] = 0.1011 to 1 >>>> this is the ratio of
NaCH3COO to CH3COOH .... so there are many combinations possible
1011 mol NaCH3COO to 1 mole CH3COOH in 1000L or 1/2 those amounts
505.5 mol NaCH3COO to 0.5 mole CH3COOH in 500L or 1/10 those
101.1 mol NaCH3COO to 0.1 mole CH3COOH in 100L or ... make it whatever you want.