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.
