Question

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.

Answer 1

See here for the background.

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....