TITRATION OF A HCl-H 3 PO 4 MIXTURE USING a ph METER Background Frequently an acid or a base is quantitatively determined by titration using ph meter to detect the equivalence point rather than using a visual indicator. This has the advantage that one actually monitors the change in ph at the equivalence point rather than just observing the change in color of a visual indicator. This eliminates any indicator blank error. Some laboratory workers complain that this method is more tedious than methods using visual indicators; they soon find, however, that after running one titration to find out the approximate location of the equivalence point, they only need to concern themselves with the drop-wise addition of titrant close to the equivalence point on subsequent titrations. The titration of a mixture of phosphoric acid and hydrochloric acid is complicated by the fact that phosphoric acid is a triprotic acid with K a1 = 7.5x10-3, K a2 = 6.2x10-8, and K a3 = 4.8x10-13. K a1 is sufficiently large that the first proton from phosphoric acid cannot be differentiated from strong acids like hydrochloric acid. The second dissociation of phosphoric acid is varies significantly from the first. The second proton can be neutralized and differentiated from the first phosphoric acid proton and the strong acid proton. Titration curves for a mixture of phosphoric acid and hydrochloric acid are illustrated here. 14 Titration of 15 meq HCl and 30 meq H 3 PO 4 with 1.0 N Base 12 10 Second Equivalence Point ph 8 6 4 First Equivalence Point 2 0 0 5 10 15 20 25 30 35 40 45 50 Titrant Volume (ml) 14
The first break in the mixed acid curve indicates the amount of hydrochloric acid plus the amount of the phosphoric acid. The amount of phosphoric acid in the sample is indicated by the difference between first and second breaks in the titration curve. The first equivalence point volume (25.0 ml) permits calculation of the total meq of HCl + (meq H 3 PO 4 )/3 since the first proton of H 3 PO 4 is neutralized. In this example 25.0 ml x 0.100 N = 2.50 meq Taking the difference between the first and second equivalence point volumes (10.0 ml) one obtains: 10.0 ml x 0.100 N x 3 eq/mole = 3.00 meq H 3 PO 4 From these two equations one can calculate that the sample contains 1.50 meq HCl and 1.00 mmol or 3.00 meq H 3 PO 4. (H 3 PO 4 has 3 equivalents per mole) This type of analysis is ideally suited for the determination of strong acid impurities in a weak acid and is unaffected by colored or suspended materials in the solution provided that these materials are not acids or bases. Interference in the analysis would be other weak or strong acids mixed into the sample. High concentrations of sodium ion or potassium ion in the sample can cause an error in the reading of the glass electrode, (i.e., the absolute ph values may be in error) but generally will not affect locating the equivalence points. Preliminary Calculations 1. How many ml of 0.100 N NaOH are required to titrate a sample containing 2.0 mmol HCl and 1.0 mmol H 3 PO 4 : (a) To the first equivalence point? and (b) To the second equivalence point? 2. Consider the titration of 20 ml of a sample that is 0.10 F in HCl and 0.05 F in H 3 PO 4 with 0.100 N NaOH. Calculate the ph of the solution being titrated at each of the following points and plot the titration curve in your notebook. Titrant Volume 0.01 ml 15.00 ml 30.00 ml 35.00 ml 40.00 ml 45.00 ml ph 15
Procedure Standardization of 0.1 N NaOH. Restandardize the 0.1 N NaOH solution prepared for the Ion Exchange experiment prior to use here. Use the same procedure as in the latter experiment. Titration of the H 3 PO 4 -HCl mixtures 1. Turn in a clean 250.0 ml volumetric flask to the laboratory instructor. Be sure that the flask is clearly marked with your name and section number. 2. Dilute the sample in the flask to the mark with boiled, distilled water. 3. Into 250 ml beakers pipet 50.00 ml portions of the acid sample. Dilute the samples with 100 ml of distilled water and stir with a magnetic stirrer. Titrate with the standard NaOH using the ph meter to detect the equivalence point.(see below) Perform an approximate titration first, adding the titrant in 0.5 ml portions. Then on subsequent titrations add the NaOH in one portion up to within 2 ml of the equivalence point. Then add the titrant in 0.10 ml, 0.05 ml, or 1 drop portions. 4. Perform the titration accurately on three portions of the acid mixture. Titrations with the ph Meter 1. Check out a ph meter, combination electrode, magnetic stirrer and stirring bar from the instructor. 2. Read the instructions on the use of the ph meter. 3. Standardize the ph meter using the buffer supplied. 4. Clean the electrode thoroughly with distilled water; drying is not necessary. 5. Immerse the electrode in the solution to be titrated; it should not go to the bottom of the titration vessel. 6. Start the stirring motor; be careful that the stirring bar does not break the glass electrode. You should allow room for the stirrer to rotate below the tip of the electrode. 7. Set the mode to ph and begin the titration. 8. Record ph and ml of titrant added. 9. Watch for the region where the ph begins to change rapidly with each added portion of titrant. As the ph begins to change more rapidly, add the titrant in smaller portions. 10. When you have passed the equivalence point by several ml, there is no reason to continue any further in the titration. 16
Treatment of the Data 1. Using good graph paper or Excel or an equivalent spreadsheet program, plot the ph against ml titrant added. Look for the regions of rapidly changing ph. 2. Also plot curves of ph/ ml as a function of ml of titrant added; the equivalence point is the point on the curve where ph/ ml has its maximum value. See the following example. 3. Duplicate titrations should agree to within 0.04 ml for excellent work. 4. Report total mmol H 3 PO 4 and mmol HCl contained in the unknown. Recall that you titrate 50 ml portions of a 250 ml total sample. Example: ml of base added ph ph ml ph/ ml 20.00 6.25 --- --- --- 20.10 6.30 0.05 0.10 0.50 20.20 6.40 0.10 0.10 1.00 20.30 6.55 0.15 0.10 1.50 20.40 6.75 0.20 0.10 2.00 20.45 6.90 0.15 0.05 3.00 20.50 7.25 0.35 0.05 7.00 20.55 8.75 1.50 0.05 30.00 20.60 9.00 0.25 0.05 5.00 20.65 9.15 0.15 0.05 3.00 20.70 9.20 0.05 0.05 1.00 20.80 9.25 0.05 0.10 0.50 (The equivalence point in this example is located at 20.55 ml) Calculations Two breaks will occur in the titration curves, the first corresponding to the titration of hydrogen ions from the HCl and the first hydrogen ion from the H 3 PO 4. The second break corresponds to the titration of H 2 PO 4 - that resulted from the H 3 PO 4. Therefore, the mmol of NaOH consumed up to the first endpoint is equal to mmol H 3 PO 4 + mmol HCl. The mmol of NaOH consumed between the first endpoint and second endpoint equals mmol H 3 PO 4. Subtract the mmol H 3 PO 4 from mmol H 3 PO 4 + mmol HCl to get mmol HCl. Multiply by the appropriate factor to get the total mmol HCl and total mmol H 3 PO 4 in 17
your 250 ml unknown sample. Report those values. Questions (required, answer these in your lab notebook) 1. What would be the effect of the presence of each of the following on locating the equivalence point for the titration with 0.100 N NaOH of a sample that is made up to 50.00 ml of 0.02 F H 3 PO 4? (a) 5.0 g NaCl (b) 5.0 g Na 2 CO 3 (c) 100 mg of an intense red azo dye (d) 5 mmol HCl 2. How will the location of the equivalence point in an acid base titration using the ph meter be affected if the meter is not calibrated properly? Why? References D. C. Harris Quantitative Chemical Analysis 8 th Ed., W. H. Freeman and Company, New York, Chapters 10 and 11 18