Suppose you are given 0.50 liter of 0.500 m acetic acid, and 0.50 liter of 0.250 liter sodium acetate. what is the maximum volume of buffer solution that you can make if the buffer must have a ph of 4.58? ka for acetic acid is 1.8 x 10-5.

First, we are going to use H-H equation :

when PH = Pka + ㏒ [acetate / acetic acid]

when we have Ka value = 1.8 x 10^-5 so, we can get the Pka value:

Pka = - ㏒Ka

       = - ㏒ 1.8 x 10^-5 

       = 4.74

by substitution:

∴ 4.58 = 4.74 + ㏒[acetate / acetic acid]

∴acetate/acetic acid = 0.692

so we need moles of acetate = volume *molarity 

                                                  = 0.5 L * 0.25 M
                                                 
                                                  = 0.125 moles

when the acetate/ acetic acid ratio = 0.692 and moles acetate = 0.125 moles

∴ moles acetic acid = 0.125 moles / 0.692 moles = 0.182 moles 

∴ volume use for 0.182 moles = 0.182 / 0.5 =  0.364 L

∴volume of buffer solution = 0.5 L + 0.364L
 
                                              = 0.864 L

Ithink that’s it’s b

Answer-ide.