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Phosphate Buffer plus NaOH and HCl

Problem

Given is a 0.03 molar phosphate buffer that consists of 4 mM KH₂PO₄ and 26 mM K₂HPO₄.

Task 1.  What is the pH of this buffer solution at 25?

Task 2.  How does the pH of the buffer solution change after

• addition of 1 mM NaOH?
• addition of 1 mM HCl?
• equilibrium with the atmospheric CO₂?

Task 3.  How does the pH of the buffer solution change after addition of 1 mM FeCl₂ and 1 mM FeCl₃? What happens when the obtained solution is oxidized by O₂?

Task 1

We start with pure water (button H2O), then click on button Reac and enter two reactants: 4 mM KH₂PO₄ and 26 mM K₂HPO₄ as shown in the right screenshot.

By clicking on the Start button, the result appears immediately:

  pH = 7.69

In words: The 0.03 molar phosphate buffer has a pH of 7.69 at 25.

Task 2

We repeat the calculation, but now we insert a third reactant into the input panel: 1 mM NaOH. The same will be done for 1 mM HCl (also as third reactant).

Finally, we set the buffer solution into the equilibrium with the atmospheric CO₂. For this purpose click on button Setup and activate the checkbox “Open CO2 System” – as shown in the right screenshot.

The obtained results for the buffer solution (in comparison with pure water) are:

buffer + 1 mM NaOH:	7.69 ⇒ 7.83		( H2O + 1 mM NaOH:	7.00 ⇒ 10.98 )
buffer + 1 mM HCl:	7.69 ⇒ 7.58		( H2O + 1 mM HCl:	7.00 ⇒ 3.01 )
buffer + atmosph. CO2:	7.69 ⇒ 7.66		( H2O + atmosph. CO2:	7.00 ⇒ 5.61 )

In fact, this example illustrates the buffer’s resistance against pH changes through acids or bases.

Task 3

The aim of the third task is to demonstrate the influence of the redox potential on mineral precipitation.

For this purpose we add to the phosphate buffer two reactants which differ in their oxidation state: 1 mM FeCl₂ (i.e., Fe(II) in the oxidation state 2) and 1 mM FeCl₃ (i.e., Fe(III) in the oxidation state 3).

The result is displayed in the right screenshot. Two pH values are given:

• Output 1 (before precipitation):  7.41
• Output 2 (after precipitation):  7.55

The mineral that precipitates is strengite (FePO₄:2H₂O). It is a Fe(III) mineral.

From the total amount of 2 mM Fe that we added to the buffer solution, 1 mM precipitates as strengite (which exactly corresponds to the amount of FeCl₃). The other 1 mM Fe remains in the solution in form of Fe(II).

Oxidation. The results above are valid for ambient redox conditions, i.e. for pe = 4.¹ Let us now change the redox conditions by oxidation with O₂ (aeration).

The corresponding input panel is shown on the right. There we have four reactants (where the first two reactants define the phosphate buffer) plus the additional condition of an Open Redox System at pe = 10.² This means that O₂ is added until pe = 10 is achieved.  (In particular, 0.25 mM O₂ is added to the system.³)

The results are shown in the right screenshot. The only difference to the previous calculation is the fact that all Fe(II) is oxidized to Fe(III). In total, there are now 2 mM Fe(III) which completely precipitates as strengite. What remains in the solution is a vanishing amount of 4.6·10^-7 mM Fe.

[Note: The input panel for reactions is designed for four reactants – see next-to-last screenshot. However, the activation of the “Open CO₂ System” and “Open Redox System” provides two additional reactants, namely CO₂ and O₂. This means that a total of 6 reactants are available.]

Remarks & Footnotes