# 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

Relevance
• david
Lv 7
1 year ago

pH 3.75 >> [H+] = 10^-3.75 = 1.78x10^-4 M

NaCH3COO --> Na+ + CH3COO- << common ion effect

CH3COOH <--> H+ + CH3COO-

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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.