Question

(a) Which two of the following compounds would you mix to make a buffer of ${}^{\mathbf{pH}}$7.45: ${}^{{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}}$(FM 98.00), ${}^{{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}}$(FM 119.98), ${}^{{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}}$(FM 141.96), and ${}^{{\mathbf{Na}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}}$(FM 163.94)?

(b) If you wanted to prepare ${}^{\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{L}}$of buffer with a total phosphate concentration of ${}^{\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{}\mathbf{M}}$, how many grams of each of the two selected compounds would you mix together?

(c) If you did what you calculated in part (b), you would not get a pH of exactly 7.45. Explain how you would really prepare this buffer in the lab.

Short answer

(a) The pH is 7.45, and the mixture contains ${}^{{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}}$and localid="1654762691970" ${}^{{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}}$finally, it arrive at the same conclusion.

(b) The two selected compound would be

$\mathbf{4}\mathbf{.}\mathbf{55}\mathbf{}{\mathbf{g}}_{\mathbf{5}}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}$

(c) The results were recorded pH via pH electrode, dilute till it's exactly what you want it to be ${}^{\mathbf{1}\mathbf{}\mathbf{L}}$with ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$.

Step-by-step solution

Definition of NaH2PO4  and Na2HPO4

• Sodium dihydrogen phosphate is a glycerol derivative made by treating edible mono and diglycerides with phosphorus pentoxide, then neutralising with sodium carbonate.
• The inorganic molecule disodium phosphate (DSP), also known as disodium hydrogen phosphate or sodium phosphate dibasic, has the formula Na2HPO4. It's a type of sodium phosphate that comes in a variety of forms.

Determine the value of pH

(a)

After the supplied mixture has been combined, the pH value must be stated whether it has changed or not.

Henderson-Hassel Balch equation

$$\begin{gathered} \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left[\mathrm{HA}\right]}{\left[H_{2}A\right]} \\ \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{2}}\mathbf{+}\mathbf{log}\frac{\left[A^2\right]}{\left[H{A}^{-}\right]} \end{gathered}$$

The pH of a solution can be determined using the formula:

role="math" localid="1654764354091" $\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}}_{\mathbf{Y}}\mathbf{}\mathbf{H}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

$$\begin{gathered} {\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{}\stackrel{{\mathbf{pK}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{148}}{\mathbf{\to }}{\mathbf{H}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}^{\mathbf{‐}}\stackrel{{\mathbf{pK}}_{\mathbf{2}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{198}}{\mathbf{\to }}{\mathbf{HPO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{‐}}\stackrel{{\mathbf{pK}}_{\mathbf{3}}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{375}}{\mathbf{\to }}{\mathbf{PO}}_{\mathbf{4}}^{\mathbf{3}\mathbf{‐}} \\ \mathbf{pH}\mathbf{\gg }\left(2.148+7.198\right)\mathbf{/}\mathbf{2} \\ \mathbf{=}\mathbf{4}\mathbf{.}\mathbf{67} \\ \mathbf{pH}\mathbf{\gg }\left(12.375+7.198\right)\mathbf{/}\mathbf{2} \\ \mathbf{=}\mathbf{9}\mathbf{.}\mathbf{79} \end{gathered}$$

The pH is 7.45, and the mixture contains${}^{{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}}$and ${}^{{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}}$finally, it arrive at the same conclusion.

Determine the weight of a compound

(b)

The mass of the chosen chemical${}^{{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}}$and ${}^{{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}}$It must be calculated.

Henderson-Hassel Balch equation

$$\begin{gathered} \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left[\mathrm{HA}\right]}{[H_{2}A\mathrm{C}]} \\ \mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{2}}\mathbf{+}\mathbf{log}\frac{\left[A^2\right]}{[H{A}^{-}]} \end{gathered}$$

The pH of a solution can be determined using the formula:

$\mathbf{pH}\mathbf{=}\mathbf{-}\mathbf{log}{\mathbf{\left[}{\mathbf{H}}^{\mathbf{+}}\mathbf{\right]}}_{\mathbf{Y}}\mathbf{}\mathbf{H}$

The negative logarithm of Hydrogen ion activity can be calculated using the equation above.

The weight of a compound can be computed as follows:

role="math" localid="1654764883235" $$\begin{gathered} \mathbf{pH}\mathbf{=}\mathbf{pK}\mathbf{2}\mathbf{+}\mathbf{log}\frac{\left[{\mathrm{HPO}}^2\right]}{\left[H_2{\mathrm{PO}}^4\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\left[{\mathrm{HPO}}_4^2\right]}{\left[H_2{\mathrm{PO}}_4^{‐}\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}{\mathbf{78}}_{\mathbf{8}} \end{gathered}$$

The final outcome can be compared.

$$\begin{gathered} \left[{\mathrm{HPO}}_4^{2‐}\right]\mathbf{+}\left[H_2{\mathrm{PO}}_4^{‐}\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{M}\mathbf{}\mathbf{gives} \\ \left[{\mathrm{HPO}}_4^{2‐}\right]\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{0320}}_{\mathbf{5}\mathbf{}}\mathbf{M} \\ \left[{\mathrm{HPO}}_4^{2‐}\right]\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{0179}}_{\mathbf{5}\mathbf{}}\mathbf{M} \\ \mathbf{}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{55}\mathbf{}{\mathbf{g}}_{\mathbf{5}}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{NaH}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}} \end{gathered}$$

Determine the exact pH of 7.45

(c)

It is necessary to explain how to make a buffer solution with a pH of 7.45.

The buffer solution with an exact pH of 7.45 can be calculated as follows:

Taking the weight off${}^{\mathbf{0}\mathbf{.}\mathbf{0500}\mathbf{}{\mathbf{molNa}}_{\mathbf{2}}{\mathbf{HPO}}_{\mathbf{4}}}$ in which it dissolves ${}^{\mathbf{0}\mathbf{.}\mathbf{9}\mathbf{}\mathbf{L}}$of a liquid Hydrochloric acid was added, and the results were recorded pH via pH electrode, dilute till it's exactly what you want it to be ${}^{\mathbf{1}\mathbf{}\mathbf{L}}$with ${}^{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}$.