How would you prepare 500 mL of a 0.01 mol/L buffer with pH 7.00 from 0.2 mol/L stock solutions of sodium dihydrogen phosphate and disodium hydrogen phosphate?

2 Answers
May 16, 2017

Answer:

See below.

Explanation:

You dilute 25 mL of each stock solution to 500 mL to get 0.01 mol/L solutions.Then you mix 300 mL of 0.01 mol/L #"NaH"_2"PO"_4# and 200 mL of 0.01 mol/L #"Na"_2"HPO"_4#.

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 = "0.01 mol/L"; color(white)(l)V_1 = "500 mL"#
#c_2= "0.2 mol/L"; color(white)(m)V_2 = ?#

#V_2 = V_1 × c_1/c_2 = "500 mL" × (0.01 color(red)(cancel(color(black)("mol/L"))))/(0.2 color(red)(cancel(color(black)("mol/L")))) = "25 mL"#

∴ You dilute 25 mL of each of the stock solutions to 500 mL 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#

or

#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.

#7 = 7.20 + log(V_("A"^"-")/V_text(HA))#

#log(V_("A"^"-")/V_text(HA)) = "7 - 7.20" = "-0.2"#

#V_("A"^"-")/V_text(HA) = 10^"-0.2" = 0.63#

(1) #V_("A"^"-") = 0.63V_text(HA)#

(2) #V_("A"^"-") + V_text(HA) =500#

Substitute (1) into (2).

#0.63V_text(HA) + V_text(HA) = 500#

#1.63V_text(HA) = 500#

#V_text(HA) = 500/1.63 = 300#

#V_("A"^"-") = "500 - 300" = 200#

Mix 300 mL of 0.01 mol/L #"NaH"_2"PO"_4# and 200 mL of 0.01 mol/L #"Na"_2"HPO"_4#.

Note: The answer can have only one significant figure, because that is all you gave for the volume, the concentrations, and the pH.

May 17, 2017

Answer:

Here's another way to prepare the buffer.

Explanation:

Add 16 mL of 0.2 mol/L #"NaH"_2"PO"_4# and 9.5 mL of 0.2 mol/L #"Na"_2"HPO"_4# to 475 mL of distilled water.

#"Moles of buffer needed" = 0.500 color(red)(cancel(color(black)("L")))× "0.01 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 mol"#

Calculate the moles of each component 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#

or

#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 moles.

#7 = 7.20 + log(n_("A"^"-")/n_text(HA))#

#log(n_("A"^"-")/n_text(HA)) = "7 - 7.20" = "-0.2"#

#n_("A"^"-")/n_text(HA) = 10^"-0.2" = 0.63#

(1) #n_("A"^"-") = 0.63Vn_text(HA)#

(2) #n_("A"^"-") + n_text(HA) =0.005#

Substitute (1) into (2).

#0.63n_text(HA) + n_text(HA) = 0.005#

#1.63n_text(HA) = 0.005#

#n_text(HA) = 0.005/1.63 = 0.0031#

#n_("A"^"-") = "0.005 - 0.0031" = 0.0019#

Calculate the volumes of each component

#V_text(HA) = 0.0031 color(red)(cancel(color(black)("mol HA"))) × "1 L HA"/(0.2 color(red)(cancel(color(black)("mol HA")))) = "0.016 L HA"#

#V_("A"^"-") = 0.0019 color(red)(cancel(color(black)("mol A"^"-"))) × (1 "L A"^"-")/(0.2 color(red)(cancel(color(black)("mol A"^"-")))) = "0.0095 L A"^"-" = "9.5 mL A"^"-"#

Add 16 mL of 0.2 mol/L #"NaH"_2"PO"_4# and 9.5 mL of 0.2 mol/L #"Na"_2"HPO"_4# to 475 mL of distilled water.

Note: I calculated the volumes to two significant figures, but only one significant figure is justified by your data.