Purpose
    The purpose of this laboratory was to gain an understanding of the natural buffer system.  We accomplished this by measuring the alkalinity of two samples from the Bushkill Stream as well as a standardized sample.

Theory
     Alkalinity is a measure of the capacity of a substance to accept protons.  In
other words, alkalinity is a measurement of buffers, or the ability of water to neutralize
a strong acid.  The three natural buffers include:  carbonate, bicarbonate, and hydroxide.

Apparatus & Reagents
- Erlenmeyer flasks
- pipettes
- burette
- pH meter
- phenolphthalein and methyl orange indicators
- 0.2 N H2SO4 (sulfuric acid)
- indicator dropper
 

Procedure
     First, we obtained 50 mL of each sample of water from the Bushkill Stream, one taken from Jacobsburg, and the other from Easton.  We also obtained 50 mL of the standard solution, sodium bicarbonate.  Each sample was placed in a separate flask.  We measured the pH of each sample using the pH meter, and also measured the conductivity.  These values are given in the table below.
          Table 1.  Initial pH and conductivity values
            Sample          Initial pH       Conductivity (ms)
            Jacobsburg        6.8                       198
            Easton               7.6                        582
            Standard            8.7                       258

For a pH which is higher than 8.3, there is a phenolphthalein titration endpoint where bicarbonate becomes the dominant carbonate form.

    The first sample we worked with was the Jacobsburg sample.  We added 5 drops of methyl orange indicator to the sample.  Using sulfuric acid as the titrant, we titrated until the endpoint was reached.  We were able to identify the approximate endpoint by the color of the solution, as it turned from a gold color to a light orange throughout the process.  We also took pH readings after every milliliter of sulfuric acid was added.  At the endpoint, the pH was 4.3.  We repeated the same process for the Easton and standard samples.  For the standard sample, we performed one additional step.  Prior to the addition of the methyl orange indicator, we added 5 drops of phenolphthalein, since its initial pH was higher than 8.3.

Results
    The following tables present the data we collected during the experiment.

Table 2.  Phenolphthalein indicator endpoint data
Sample        Initial pH      Amt. Of Phen.        Amt. Of Titrant       pH at Endpoint
Jacobsburg     6.8              0 drops                     0 mL                          n/a
Easton            7.6               0 drops                    0 mL                           n/a
Standard         8.7              5 drops                    0.7 mL                        7.1

Table 3.  Methyl orange endpoint data
Sample          pH      Amt. Of Meth. Orange    Amt. Of Titrant      pH at Endpoint
Jacobsburg   6.8                5 drops                     2.8 mL                       4.3
Easton          7.6                5 drops                    11.6 mL                      4.3
Standard       7.1                5 drops                     7.5 mL                       4.3

    Table 2 indicates that we added phenolphthalein only to the standard sample in addition to the methyl orange, because its initial pH was greater than 8.3.  We titrated drop by drop until the pink phenolphthalein solution turned clear.  We measured the pH of that point, and entered it into the table.
    Table 3 indicates that we added only methyl orange to the Jacobsburg and Easton samples.  Phenolphthalein was not necessary because the initial pH values for these samples was below 8.3.  We titrated all samples until the pH reading reached 4.3.  Table 3 also presents the amount of titrant (sulfuric acid) necessary to reach this point.

Calculations

Alk (eq/L)*Volsample = Nacid (eq/L)*Volacid

Alk (mg/L as CaCO3) = Alk (eq/L)*50,000 mg CaCO3/eq

Phenolphthalein alkalinity (as mg/L of CaCO3) =
Total alkalinity (as mg/L of CaCO3) =
Where:      A = mL of acid to reach Phenolphthalein end point
                 B = mL of acid to reach Methyl Orange end point
                 N = Normality of acid
 
 
 
 

Jacobsburg:
Alk (eq/L) * Volume of sample = 0.02 eq/L * 2.8 mL *  __1 L_  =  5.6 x 10^-5 eq/L
                                                                                    1000mL
As CaCO3 = 5.6 x 10^-5 eq/L *  50,000 mg CaCO3 / eq  =  2.8 mg CaCO3 / L

Total Alk. = 2.8 mL x .02 N x 50,000  = 56 mg/L as CaCO3
                                    50

Easton:

Alk (eq/L) * Volume of sample = 0.02 eq/L * 11.6 mL * __1L__ =  2.32 x10^-4 eq/L
                                                                                     1000 mL

As CaCO3 = 2.32 x 10^-4 eq/L * 50,000 mg CaCO3/eq = 11.6 mg as CaCO3/L

Total Alk. = 11.6 mL x .02 N x 50,000  = 232 mg/L as CaCO3
                                   50 mL

Standard:

Phenolphthalein Alk. =  0.7 mL x 0.02 eq/L x 50,000  =  14 mg as CaCO3/L
As CaCO3                                      50 mL

Total Alk.    = 7.5 mL x 0.02 eq/L x 50,000 =  150 mg/L as CaCO3
As CaCO3                    50 mL
 
 

Conclusion
    Based on the data from Table 3, we concluded that Easton the highest alkalinity, as it required the most titrant to reach a pH of 4.3.  We can attribute this to a higher amount of limestone in the Easton area, as compared to Jacobsburg.

Discussion
1. Why is the value of 50,000 in each of the alkalinity equations?

The 50,000 is needed to convert alkalinity measured in equivalence per liter to milligrams per liter as calcium carbonate.

2. Did anything other than hydroxide, carbonate, and bicarbonate contribute to alkalinity?  If so, what and how?

Yes, anything that can accept hydrogen protons can contribute to alkalinity.  Therefore, any form of contamination can alter alkalinity.

3. If you had a body of pure water in a closed system then introduced an
atmosphere of oxygen and carbon dioxide, would you be surprised if the pure water began to show a concentration alkalinity?  If no, explain why.

No, we would not be surprised because carbon dioxide and water forms carbonic acid.  When carbonic acid dissociates, it forms bicarbonate, which is one of the natural buffers, and contributes to alkalinity.

4. Relate conductivity with alkalinity.  Does it make sense that conductivity fluctuates with alkalinity?  Explain.

Conductivity is a result of total dissolved ions (TDS) in solution.  High conductivity may indicate contamination with salts or inorganic wastes.  Alkalinity is a measure of buffers, so when conductivity increases, alkalinity may increase as well, but not necessarily.  In Laboratory 4, this was true, as conductivity increased, alkalinity increased.  This may not always occur.  Ions in a body of water may be present that are not included in a buffer system, resulting in increased conductivity without increased alkalinity.

5. Why does the phenolphthalein turn clear immediately in all the Bushkill Stream samples?  Explain in a clear and precise manner.

Phenolphthalein turns clear immediately in the Bushkill Stream samples because the turning point of phenolphthalein is 8.3.  Above 8.3, solutions with phenolphthalein turn pink.  Below 8.3, the solution is clear.  Since the pH of the river samples were below 8.3, the phenolphthalein will not turn pink.

6. Explain whether or not your results make sense based on what you know
about the watershed geology.  Which part of the stream is the better buffer
against acid precipitation?

Yes, our results make sense, because the limestone found in the Easton area adds to the concentration of bicarbonate in the water.  There is a higher alkalinity is this region as a result.  Because of this, the Easton region is a better buffer against acid precipitation.

7. The 0.02N solution of sulfuric acid was prepared ahead of time by diluting
2.8 mL of concentrated sulfuric acid (r=1.835 g/cm3) into 1000 mL of deionized water, and diluting the solution by a factor of 5.  Show that this procedure results in a 0.02N solution of sulfuric acid.

      0.02N sulfuric acid:

      2.8 mL * 1.835 g     =  5.138 g sulfuric acid / 1 L H20
                       1 mL
     5.138 g sulfuric acid  * 1 mol sulfuric acid   *  2  eq sulfuric acid  =  0.1eq/L
             1 L H20                  98 g sulfuric acid                 1 mol

      0.1 eq/L  =   0.02N
           5