Question

Question #b6a38

Answer 1

Start by figuring out what the concentrations the components of the buffer will have in the total solution.

The total volume of the buffer will be

`V_("total") = "250 mL" + "250 mL" = "500 mL"`

The number of moles of acetic acid you'll use for the buffer is

`C = n/V => n = C * V = 250*10^(-3)"L" * "0.0100 M" = 2.5 * 10^(-3)"moles"`

This means that the concentration of acetic acid in the buffer will be

`C_("buffer") = n/V_("total") = (2.5 * 10^(-3)"moles")/(500 * 10^(-3)"L") = "0.005 M"`

Now do the same for sodium acetate

`n = C * V = 250 * 10^(-3)"L" * "0.100 M" = 25 * 10^(-3)"moles"`

This means that the concentration of sodium acetate in the buffer will be

`C_("buffer") = n/V_("total") = (25 * 10^(-3)"moles")/(500 * 10^(-3)"L") = "0.05 M"`

All you have to do now is use the Henderson-Hasselbalch equation to determine the pH of the buffer

`[pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept) = pKa + log(([CH_3COONa])/([CH_3COOH]))`

`pH = 4.76 + log(("0.05 M")/("0.005 M")) = 4.76 + 1 = 5.76`