Phosphate Buffer plus NaOH and HCl


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

Task 3.  How does the pH of the buffer solution change after addition of 1 mM FeCl2 and 1 mM FeCl3?  And, what happens when the Fe-containing solution is oxidized by O2?

Task 1

phosphate buffer input

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. Then, the same should be done for 1 mM HCl (also as third reactant).

phosphate buffer plus CO2

Finally, we set the buffer solution into the equilibrium with the atmospheric CO2 (without the third reactant NaOH or HCl).

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 )

These examples illustrate 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).

phosphate buffer plus FeCl2 and FeCl3

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

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).

phosphate buffer plus FeCl2 and FeCl3 at pe 10

Oxidation.  The above results 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 are 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)

pH of phosphate buffer plus FeCl2 and FeCl3 at pe 10

The result is 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 — as can be seen in the next-to-last screenshot. However, the “Open CO2 System” and “Open Redox System” provide two additional reactants, namely CO2 and O2. This makes a total of 6 reactants available.]

Remarks & Footnotes

  1. pe = 4 is the default parameter for the redox potential in PhreeqC. 

  2. The results are almost independent of the exact value of pe, when it is high enough. Thus, you can choose any value pe > 8 to characterize an oxidized aquatic environment. 

  3. The value of 0.25 mM O2 is presented in the program’s output tables. 

[last modified: 2023-11-15]