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?

May 16, 2017

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 ${\text{NaH"_2"PO}}_{4}$ and 200 mL of 0.01 mol/L ${\text{Na"_2"HPO}}_{4}$.

Calculate the dilution of the stock solutions

We can use the dilution formula

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${c}_{1} = \text{0.01 mol/L"; color(white)(l)V_1 = "500 mL}$
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

$\text{H"_2"PO"_4^"-" + "H"_2"O" → "H"_3"O"^"+" + "HPO"_4^"2-"; "p} {K}_{\textrm{a}} = 7.20$

or

$\textcolor{w h i t e}{m} \text{HA" color(white)(ll)+ "H"_2"O" → "H"_3"O"^"+" + color(white)(m)"A"^"-}$

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.

$7 = 7.20 + \log \left({V}_{\text{A"^"-}} / {V}_{\textrm{H A}}\right)$

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

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

(1) ${V}_{\text{A"^"-}} = 0.63 {V}_{\textrm{H A}}$

(2) ${V}_{\text{A"^"-}} + {V}_{\textrm{H A}} = 500$

Substitute (1) into (2).

$0.63 {V}_{\textrm{H A}} + {V}_{\textrm{H A}} = 500$

$1.63 {V}_{\textrm{H A}} = 500$

${V}_{\textrm{H A}} = \frac{500}{1.63} = 300$

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

Mix 300 mL of 0.01 mol/L ${\text{NaH"_2"PO}}_{4}$ and 200 mL of 0.01 mol/L ${\text{Na"_2"HPO}}_{4}$.

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

Here's another way to prepare the buffer.

Explanation:

Add 16 mL of 0.2 mol/L ${\text{NaH"_2"PO}}_{4}$ and 9.5 mL of 0.2 mol/L ${\text{Na"_2"HPO}}_{4}$ to 475 mL of distilled water.

$\text{Moles of buffer needed" = 0.500 color(red)(cancel(color(black)("L")))× "0.01 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.005 mol}$

Calculate the moles of each component needed

The chemical equation for the buffer is

$\text{H"_2"PO"_4^"-" + "H"_2"O" → "H"_3"O"^"+" + "HPO"_4^"2-"; "p} {K}_{\textrm{a}} = 7.20$

or

$\textcolor{w h i t e}{m} \text{HA" color(white)(ll)+ "H"_2"O" → "H"_3"O"^"+" + color(white)(m)"A"^"-}$

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.

$7 = 7.20 + \log \left({n}_{\text{A"^"-}} / {n}_{\textrm{H A}}\right)$

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

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

(1) ${n}_{\text{A"^"-}} = 0.63 V {n}_{\textrm{H A}}$

(2) ${n}_{\text{A"^"-}} + {n}_{\textrm{H A}} = 0.005$

Substitute (1) into (2).

$0.63 {n}_{\textrm{H A}} + {n}_{\textrm{H A}} = 0.005$

$1.63 {n}_{\textrm{H A}} = 0.005$

${n}_{\textrm{H A}} = \frac{0.005}{1.63} = 0.0031$

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 ${\text{NaH"_2"PO}}_{4}$ and 9.5 mL of 0.2 mol/L ${\text{Na"_2"HPO}}_{4}$ 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.