Calculate the dilution of the stock solutions
We can use the dilution formula
#color(blue)(|bar(ul(color(white)(a/a)c_1V_1 = c_2V_ 2color(white)(a/a)|)))" "#
#c_1 = "0.2 mol/L"; color(white)(ll)V_1 = "50 mL"#
#c_2= "0.01 mol/L"; V_2= ?#
#V_2 = V_1 × c_1/c_2 = "50 mL" × (0.2 color(red)(cancel(color(black)("mol/L"))))/(0.01 color(red)(cancel(color(black)("mol/L")))) = "1000 mL" = "1 L"#
∴ We dilute 50 mL of each of the stock solutions to 1 L to prepare the 0.01 mol/L solutions for the buffer.
Calculate the volumes of each solution needed
The chemical equation for the buffer is
#"H"_2"PO"_4^"-" + "H"_2"O" → "H"_3"O"^"+" + "HPO"_4^"2-"; "p"K_text(a) = 7.20#
#color(white)(m)"HA" color(white)(ll)+ "H"_2"O" → "H"_3"O"^"+" + color(white)(m)"A"^"#
The Henderson-Hasselbalch equation is
#"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))#
Both solutions have the same concentration, so the ratio of the volumes is the same as the ratio of the molarities.
#6.6 = 7.20 + log(V_("A"^"-")/V_text(HA))#
#log(V_("A"^"-")/V_text(HA)) = 6.6 - 7.20 = "-0.6"#
#V_("A"^"-")/V_text(HA) = 10^"-0.6" = 0.25#
(1) #V_("A"^"-") = 0.25V_text(HA)#
(2) #V_("A"^"-") + V_text(HA) = 100#
Substitute (1) into (2).
#0.25V_text(HA) + V_text(HA) = 100#
#1.25V_text(HA) = 100#
#V_text(HA) = 100/1.25 = 80#
#V_("A"^"-") = "100 - 80" = 20#
Use 80 mL of 0.01 mol/L #"NaH"_2"PO"_4# and 20 mL of #"NaH"_2"PO"_4#.
