# Question 97e59

Apr 14, 2017

#### Answer:

You need 80 mL of 0.01 mol/L ${\text{NaH"_2"PO}}_{4}$ and 20 mL of ${\text{Na"_2"HPO}}_{4}$.

#### Explanation:

Calculate the dilution of the stock solutions

We can use the dilution formula

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

${c}_{1} = \text{0.2 mol/L"; color(white)(ll)V_1 = "50 mL}$
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

$\text{H"_2"PO"_4^"-" + "H"_2"O" → "H"_3"O"^"+" + "HPO"_4^"2-"; "p} {K}_{\textrm{a}} = 7.20$
$\textcolor{w h i t e}{m} \text{HA" color(white)(ll)+ "H"_2"O" → "H"_3"O"^"+" + color(white)(m)"A"^}$

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.

$6.6 = 7.20 + \log \left({V}_{\text{A"^"-}} / {V}_{\textrm{H A}}\right)$

log(V_("A"^"-")/V_text(HA)) = 6.6 - 7.20 = "-0.6"

V_("A"^"-")/V_text(HA) = 10^"-0.6" = 0.25

(1) ${V}_{\text{A"^"-}} = 0.25 {V}_{\textrm{H A}}$

(2) ${V}_{\text{A"^"-}} + {V}_{\textrm{H A}} = 100$

Substitute (1) into (2).

$0.25 {V}_{\textrm{H A}} + {V}_{\textrm{H A}} = 100$

$1.25 {V}_{\textrm{H A}} = 100$

${V}_{\textrm{H A}} = \frac{100}{1.25} = 80$

V_("A"^"-") = "100 - 80" = 20#

Use 80 mL of 0.01 mol/L ${\text{NaH"_2"PO}}_{4}$ and 20 mL of ${\text{NaH"_2"PO}}_{4}$.