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Chemistry LibreTexts

Sample Pretreatment

  • Page ID
    220395
  • Q1. Under what conditions should the water samples be stored and within what time frame should the samples be analyzed?

    Students should be encouraged to research guidelines for sample storage and time frame within which analyses need to be completed before the sample degrades.  They will soon realize that conditions vary dramatically depending on the analyte.  For example, metals solutions, once acidified, are stable for several months, while, as in the case of nitrate, analysis needs to be completed immediately if possible but no later than 48 hours.  The table included in the module may be a good starting point for a discussion.

    Q2. What changes could take place that would cause sample degradation?

    Many physical and chemical changes can occur if a sample is not stored properly.  Temperature, light exposure or pH conditions may contribute to sample degradation.  For example, metals can precipitate as hydroxides if the sample is not acidified.  Ammonia can be lost from samples that are not refrigerated or acidified.

    Q3.  No preservation method is required for nitrates.  However, a water sample can only be stored for up to 48 hours.  What possible degradation could occur over an extended period of time?  What chemical preservative could be added to prolong the life of the sample?  How could you devise an experiment to test whether compositional changes for nitrates occur over time?

    Natural waters contain bacteria that will use nitrates as nitrogen source.  Refrigeration will slow down nitrate degradation and acidification of the sample also prolongs the sample over a longer period of time.  Students can be guided to design an experiment where the concentration of nitrates is measured at regular intervals of time to determine the degradation rate.  Students should also consider that such rate will vary depending on the specific bacteria population and other parameters such was water temperature and pH.

    Q4.  Cr(VI) must be analyzed within 24 hours.  What might happen to compromise the analysis of Cr(VI) if the samples were held only longer before completing the analysis?

    Chromium (VI) is a strong oxidizing agent.  That means that it will readily react with any species and be reduced to chromium (V) or chromium (III).

     Q5.  Preservation of water samples for metals analysis requires acidification with nitric acid below pH 2.  Why is this procedure required for analysis of calcium and magnesium? 

    Calcium and magnesium will form hydroxides.  In the next question students are asked to compare the solubility of Ca(OH)2 and Mg(OH)2 at pH 2 and pH 8.  They may be further asked to evaluate an ideal pH value where the solubility becomes large enough to not pose a problem of potential precipitation.

    Q6.  Compare the solubility of Ca(OH)2 and Mg(OH)2 at pH 2 and pH 8.  What could happen to the Ca2+ and Mg2+ ions stored over time if the pH was not adjusted to an acidic value?

    The solubility equilibrium established in a saturated solution of Ca(OH)2 is:

    \[\ce{Ca(OH)2 (s)  ↔ Ca^2+ (aq)  +  2 OH- (aq)}\nonumber\]

    The equilibrium constant expression for Ca(OH)2 is:

    \[\mathrm{K_{sp} = [Ca^{2+}] [OH^-]^2}\nonumber\]

    The molar solubility of Ca(OH)2 can be defined as [Ca2+] because dissolving one mole of Ca(OH)2 provides one mole of calcium ions. Solving for the molar solubility of Ca(OH)2 may be accomplished using the following steps.

    For pH = 2

    \[\mathrm{[H^+] = 10^{-pH} = 10^{-2} = 0.01\: M}\nonumber\]

    \[\mathrm{[OH^-] = \dfrac{K_w}{[H^+]}   =  \dfrac{1.0 \times 10^{-14}}{0.01\: M} = 1.0 \times 10^{-12}\:  M}\nonumber\]

    \[\mathrm{[Ca^{2+}] = \dfrac{K_{sp}}{[OH^-]^2} =  \dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-12}\: M)^2} = 6.5 \times 10^{18}\:  M}\nonumber\]

    For pH = 8

    \[\mathrm{[H^+] = 10^{-pH} = 10^{-8} = 1.0 \times 10^{-8}\: M}\nonumber\]

    \[\mathrm{[OH^-] = \dfrac{K_w}{[H^+]} = \dfrac{1.0 \times 10^{-14}}{(1.0 \times 10^{-8}\: M)} = 1.0 \times 10^{-6}\:  M}\nonumber\]

    \[\mathrm{[Ca^{2+}] = \dfrac{K_{sp}}{[OH^-]^2} = \dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-6}\: M)^2} = 6.5 \times 10^6\:  M}\nonumber\]

    Similar calculations can be carried out for magnesium.

    The table below summarizes the solubilities of calcium and magnesium hydroxides at pH 2 and 8.

     

     

    Molar Solubility (M)

    Salt

    Ksp

    pH 8

    pH 2

    Ca(OH)2

    6.5 x 10-6

    6.5 x 102

    6.5 x 1018

    Mg(OH)2

    7.1 x 10-12

    7.1

    7.1 x 1012

    One can observe that the solubility of metal hydroxides increases with decreasing pH. In fact, metal hydroxides that have a cation with an oxidation state of +2 exhibit a one hundred-fold increase in their solubility for every unit decrease in pH. Metal hydroxides that have a cation with an oxidation state of +3 exhibit a one thousand-fold increase in their solubility for every unit decrease in pH.  Therefore, acidification of the samples is needed to keep calcium and magnesium dissolved in water samples from precipitating over times as hydroxides, thus decreasing the actual amount of these ions and providing an inaccurate estimate of their concentration.

    Note that for salts having a relatively large Ksp or under acidic conditions, the predicted molar solubility and mass solubility are unrealistically large. In these cases, the solubility is no longer dependent on the solubility equilibrium but on the number of moles of solvent required to interact with each mole of solute. The molarity of water is 55.6 M, and each ion would be surrounded by several molecules of water. It is, therefore, unreasonable to expect that the solute concentration under low pH conditions would exceed several moles per liter.

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