Acidification with Strong Acids to pH 3

Problem

Given is a groundwater sample with pH 6.9. How much nitric acid (HNO3) and how much sulfuric acid (H2SO4) are needed to obtain pH 3?

Acidification with HNO3 (Monoprotic Acid)

Click on Open and select the groundwater sample gw.sol. The only thing we have to do is to change the pH from 6.9 to 3.0. Then, click on Start. Immediately you will get the message that the CBE is 81.9%.

To adjust charge balance select from the parameter list the parameter N(5)1, and click on next. The output is:

ΔN(5) = 6.93 mM

In words, we need 6.93 mM of the strong acid HNO3 in order to decrease the pH from 6.9 to exactly 3.0. The message window tells us also that the nitrate concentration increases from 2.6 auf 426 mg/L.

Acidification with H2SO4 (Diprotic Acid)

Repeat the above procedure, but now select the parameter S(6)2 to adjust charge balance. The result is:

ΔS(6) = 3.55 mM

This is a little bit more than the half of the HNO3 amount. (It is not exactly the half amount because sulfuric acid is diprotic, but only the first dissociation step belongs to strong acids.)

From the output tables (table Ions) you can extract the following data for the solution before and after the acidification (concentration units in mM):

pH SO4-2 HSO4- sulfate complexes total sulfate S(6)
6.9 0.175 0.010 0.046 0.185
3.0 3.045 0.164 0.575 3.209

The column “sulfate complexes” displays the sum of more than 10 sulfate-complex species listed in table Ions. The most prevalent sulfate complexes in the given water sample are the uncharged complexes CaSO4 and MgSO4.

The total sulfate S(6) in the last column is the sum of:

S(6)   =   SO4-2 + HSO4- + sulfate complexes

1 The symbol N(5) abbreviates nitrogen in the oxidation state +5 which comprises all nitrate species including NO3-.

2 The symbol S(6) abbreviates sulfur in the oxidation state +6 which comprises all sulfate species including SO4-2 and HSO4-.

[last modified: 2013-08-05]