Why is one a buffer solution and not the other?

A student prepares two solutions:
Solution A is prepared by mixing #50cm^3# of #0.100mol# #dm^(-3)# #CH_3COOH_((aq))# with #25cm^3# of #0.100mol# #dm^(-3)# #NaOH_((aq))#
Solution B is prepared by mixing #25cm^3# of #0.200mol# #dm^(-3)# #CH_3COOH_((aq))# with #50cm^3# of #0.100mol# #dm^(-3)# #NaOH_((aq))#

Explain why solution A is a buffer solution, but solution B is not.

1 Answer
anor277
Dec 4, 2017

Explanation:

A buffer equation is a an equilibrium mixture of a weak base and its conjugate acid (or vice versa) in APPRECIABLE concentrations, that acts to resist GROSS changes in #pH#.

For #"solution A"# we gots #"acetic acid"# and #"sodium acetate"# on a 2:1 molar ratio....the pH of this buffer will be reduced from the #pK_a# of #"acetic acid"=4.76#

For #"solution B"# we adds a TWO molar excess of sodium hydroxide to #"acetic acid"# and thus the solution is stoichiometric in #"sodium acetate"#....and #"sodium hydroxide"#. There is a substantial concentration of #NaOH(aq)#, which expresses itself as a high #pH#.

For #"solution A"# using the equation given in the link you could could find the #pH# of the solution....

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