The answer is #"23.0 mL"# of disodium phosphate solution is needed for this particular buffer.

So, you know that the total volume of the buffer will be #"40.0 mL"#, or #"0.04 L"#. This means that the sum of the volumes of the two solutions must be

#V_1 + V_2 = 0.04# **(1)**

SInce you're dealing with a buffer, use the **Henderson-Hasselbalch equation** to determine another relationship between #V_1# and #V_2#

#[pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept) = pKa + log(([Na_2HPO_4])/([NaH_2PO_4]))#

#6.95 = 6.82 + log(([Na_2HPO_4])/([NaH_2PO_4])) => ([Na_2HPO_4])/([NaH_2PO_4]) = 1.35# **(2)**

Now, use the initial concentrations of the two solutions to find the relationship you need. For sodium phosphate, the buffer concentration can be written as

#[NaH_2PO_4] = n_("sodium")/(V_1+V_2)#

The number of moles of sodium phosphate is

#n_("sodium") = C * V_1 = "0.100 M" * V_1 = 0.100 * V_1#

Likewise, the concentration of disodium phophate is

#[Na_2HPO_4] = n_("disodium")/(V_1+V_2)#

Once again, the number of moles can be calculated using the initial concentration

#n_("disodium") = C * V_2 = "0.100 M" * V_2 = 0.100 * V_2#

Plug all of this into equation **(2)**

#(0.100 * V_2)/(V_1 + V_2) * (V_1 + V_2)/(0.100 * V_1) = 1.35#, or

#V_2/V_1 = 1.35 => V_2 = 1.35 * V_1#. Plug this into equation **(1)**

#V_1 + 1.35 * V_1 = 0.04 => V_1 = 0.04/2.35 = 0.017#, which means that

#V_2 = 0.04 - 0.017 = 0.023#

Therefore, the volume of disodium phosphate #"0.100 M"# solution you need for this particular buffer is

#V_2 = "0.023 L"= "23.0 mL"#