Question

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?

Answer 1

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.

Answer 2

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.