By utilizing acid-base titration. we determined the suitableness of phenolphthalein and methyl ruddy as acerb base indexs. We found that the equality point of the titration of hydrochloric acid with Na hydrated oxide was non within the ph scope of phenolphthalein’s colour scope. The titration of acetic acid with Na hydrated oxide resulted in an equality point out of the scope of methyl ruddy. And the titration of ammonium hydroxide with hydrochloric acid had an equality point that was besides out of the scope of phenolphthalein. . The methyl ruddy index and the phenolphthalein index were unsuitable because their pH ranges for their colour alterations did non cover the equality points of the tests in which they were used. However. the methyl ruddy index is more suited. since it’s pH scope is closer to the equality points of the titrations.
Acid-base reactions are one of the most common and of import chemical interactions. They are critical to both environmental and industrial systems. As an of import variable. pH controls the toxicity. mobility. solubility. and destiny of many aquatic ecosystems. Most aquatic life signifiers can non last outside a pH window from about 4. 5 to 9. From an industrial point of view. use of pH is both a tool for and a requirement to all H2O intervention processes. 1 Along with pH indexs. titration is a critical tool in finding the factors of many commercial and environmental systems.
Therefore. cognition of acid-base titration curves is critical to the environmental scientist. Titration. an analytical technique. allows the quantitative finding of a dissolved substance being titrated. known as an analyte. Titration requires cognition of the equality point: a theoretical point where the chemical equivalents of titrant added are precisely equal to the chemical equivalents of the solute being titrated. It besides requires the cognition of the End point: an operational point which approximates the place of the equality point by some physical alteration. 1
The intent of this lab is to happen the most suited index. In the lab we used. Phenolphthalein or Methyl-Red. for differing titrations. Here are some other indexs: 1
In this lab we used the undermentioned reactions and equations in this lab:
Chemical reaction 1: The neutralisation reaction of Sodium Hydroxide and Hydrochloric Acid:
HCl ( aq ) + NaOH ( aq ) ? NaCl ( aq ) + H2O ( cubic decimeter )
Chemical reaction 2: The complete dissociation reaction of Hydrochloric Acid:
HCl ( aq ) + H2O ( cubic decimeter ) ? H3O+ ( aq ) + Cl- ( aq )
Chemical reaction 3: The uncomplete dissociation reaction of Acetic Acid:
CH3COOH ( aq ) + H2O ( cubic decimeter ) ? H3O+ ( aq ) + CH3COO- ( aq )
Chemical reaction 4: The reaction of the Acetate Ion with Water:
CH3COO- ( aq ) + H2O ( cubic decimeter ) ? CH3COOH ( aq ) + H3O+ ( aq )
Chemical reaction 5: The reaction of Ammonia with Hydrochloric Acid:
NH3 ( aq ) + HCL ( aq ) ? NH4+ ( aq ) + Cl- ( aq )
Chemical reaction 6: The reaction of the left over Ammonia with Water:
NH4+ ( aq ) + H2O ( cubic decimeter ) ? NH3 ( aq ) + H3O+ ( aq )
The three following mathematical equations were used:
Equation 1: To cipher the mass of a substance needed to fix a solution:
Mass of substance ( in gm ) = ( M ) ( V ) ( Formula Weight )
Equation 2: To cipher the pH:
pH = -log [ H3O+ ]
Equation 3: To cipher the volume of titrant added to the analyte to bring forth a concluding solution with a specific Hydronium concentration:
[ H3O+ ] = ( Vanalyte x Manalyte ) – ( Vtitrant ) ( Mtitrant ) / ( Vtitrant + Vanalyte )
Equation 4: To cipher the figure of moles:
Number of Moles = Molar Concentration Volume = Mass Molar Mass
-0. 1M NaOH ( prepared from solid NaOH )
-0. 1M CH3COOH ( aq )
-0. 1M Hcl ( aq )
-0. 1M NH3 ( aq )
-Methyl-red index ( 1 ( 1
-electronic pH metre
-50mL burette with base
-25mL volumetric pipette
-pipette safety bulb
-magnetic splash saloon
-plastic wash bottle
-500mL volumetric flask and stopper
-plastic or glass funnel
Followed the processs listed in the University of Winnipeg Laboratory Manual was followed. 2
Alterations: Alternatively of thining the titrant with 75mL of H2O. we diluted it with 50mL of H2O.
Part I: Datas and Calculations
Table 2: Titration Data Table
Trial 1 – HCl and NaOH ( Strong-Acid + Strong-Base )
Trial 2 – CH3COOH and NaOH ( Weak-Acid + Strong-Base )
Trial 3 – NH3 and HCl ( Weak-Base + Strong-Acid )
( * ) marks equality scope
Trial 1Trial 2Trial 3
Titrant Volume ( milliliter )
pHTitrant Volume ( milliliter ) pHTitrant Volume ( milliliter ) pH
0. 001. 70. 003. 10. 0010. 4
5. 001. 75. 004. 05. 009. 7
10. 001. 710. 004. 410. 009. 3
15. 001. 815. 004. 815. 008. 9
20. 002. 020. 005. 120. 007. 9*
21. 002. 121. 005. 321. 006. 0
22. 002. 122. 005. 421. 503. 8
23. 002. 223. 005. 522. 003. 1
24. 002. 224. 005. 722. 502. 9
25. 002. 325. 006. 2*23. 002. 7
26. 002. 425. 506. 723. 502. 6*
27. 002. 726. 009. 824. 002. 5
28. 003. 026. 5010. 524. 502. 4
29. 003. 3*27. 0011. 2*25. 002. 4
30. 003. 827. 5011. 426. 002. 3
30. 509. 428. 0011. 627. 002. 2
31. 0010. 3*29. 0011. 728. 002. 2
31. 5010. 830. 0011. 830. 002. 1
32. 0011. 031. 0011. 835. 002. 0
33. 0011. 232. 0011. 840. 001. 9
35. 0011. 534. 0011. 9
40. 0011. 835. 0011. 9
40. 0012. 0
Table 3: Equality Point values.
Titration of HCl with NaOH7. 0
Titration of CH3COOH with NaOH8. 6
Titration of NH3 with HCl 5. 5
1 ) Initial pH
See Table 2.
2 ) Final pH
See Table 2.
3 ) Equality Scope
Using Graph 1: The Volume of Titrant Added in order to make the Endpoint and the Corresponding pH Values. detect the perpendicular line of each titration and see the points in which the horizontal lines intersect it. These points give the Equivalence Range. See Table 2 for values.
4 ) Equality Point
Using Graph 1: The Volume of Titrant Added in order to make the Endpoint and the Corresponding pH Values a perpendicular line from the center of the perpendicular equality scope is drawn so that it will cross the y-axis. The point at which it intersects the y-axis is the pH of the equality point. See Table 3 for values.
5 ) Calculation of the Mass Needed to Form 0. 1 M NaOH
Find Number of Moles of NaOH needed
0. 1M 0. 5 L = 0. 05 mol NaOH
Find Amount of NaOH needed in Grams
0. 05 mol NaOH mol mass NaOH = gms of NaOH needed
0. 05 mol ( 22. 989770 + 15. 9994 + 1. 00794 ) = 1. 999g NaOH
Part II: Observations
Table 4: Observations of color alterations of Acid-Base indexs during Titration.
Initial coloring material ( before alteration ) End point colourpH at End Point ( color alteration )
Titration of HCl with NaOHclearPinky-red10. 1
Titration of CH3COOH with NaOHredyellow6. 0
Titration of NH3 with HCl Pinkyclear8. 9
The finding of the terminal points and equality points of assorted titrations allow for an rating of how suited an index is for a peculiar reaction. The titration of hydrochloric acid with Na hydrated oxide resulted in a impersonal equality point at a pH of 7. 0. Phenolphthalein. the acid-base index that was used in this titration. has a colour alteration scope of pH scope of 8. 2 to 9. 8 ( see Table 1 ) and produced an end point coloring material alteration with Trial at a pH of 10. 1. This index is unsuitable since it does non cover the scope of the equality point which is 7. 0.
The titration of acetic acid with Na hydrated oxide resulted with a somewhat basic equality point at a pH of 8. 6. The methyl ruddy index used in this titration was non suited in the titration of hydrochloric acid by Na hydrated oxide because its equality point is out of methyl red’s scope ( pH 4. 4 to 6. 2 from Table 1 )
The usage of phenolphthalein as the acid-base index in the titration of ammonium hydroxide with hydrochloric acid was unsuitable since the scope of
phenolthailen ( pH8. 2 to 9. 8 ) does non cover the equality point. which is pH of 5. 6.
Therefore. none of the indexs were suited for the titrations they were involved with.
1 ) Based on your consequences. which index of the two is more suited for this type of titration experiment? Explain.
Methyl Red is the better index. The equality points for the titrations fell more closely with Methyl Red’s scope of colour alteration ( pH of 4. 4to 6. 2 ) .
2 ) Calculate the volume of 0. 1 M NaOH which must be added to 50 milliliter of 0. 1 M HCl to give a concluding solution of pH 6.
To happen [ H+ ] . Take the reverse log:
10-6 = [ H+ ] = 1 10-6
Find the entire figure of initial moles of HCl
Entire Initial moles = 0. 1 M HCl 0. 050 L = 0. 0050 mol HCl
Since our the end pH is 6 ( acidic ) acid must be left over. Therefore the volume of NaOH added is unknown ( omega )
Mol NaOH = 0. 1 M ( omega )
The entire figure of concluding moles of HCl:
Mol HCl concluding = moles of HCl initial – moles of HCl reacted ( aka mols NaOH )
Mol HCl concluding = 0. 0050 mol HCl – ( 0. 1M ) ( omega )
The entire volume:
Entire Volume = initial volume of HCl + volume of NaOH added
Entire Volume = 0. 050 L + omega
The concentration of HCl:
[ H+ ] = mol H+ / entire volume
1. 6 10-6 = ( 0. 0050 – 0. 1z ) / ( 0. 050 + omega )
0. 0050 – 0. 1z = 5. 0 10-8 + 1. 0 10-6
omega = 0. 049999L
Therefore 49. 99mL of NaOH must be added.
Beginnings of Mistake
There are legion possible beginnings of mistake. The largest beginning of mistake in this experiment involves adding the titrant. Although the processs indicate to add 5. 00 milliliter. 1. 00 milliliter. and 0. 50 mL sums of titrant as the equality point is near to being reached. 0. 5 milliliter of titrant is excessively big to be added. Adding 0. 5mL may “overshoot” the end point. doing the consequences inaccurate. A possible solution is to utilize 0. 10 mL increases alternatively of 0. 50 milliliter. Yet. due to the restrictions of human accomplishment at adding the titrant. it may be difficult to accomplish smaller sums and more accurate consequences.
Other beginnings of mistake include human mistakes. For illustration. taint of samples may interfere with the titration’s consequences. A solution to this job is taking attention when adding. blending. or cleansing.
By utilizing acid-base titration. we found that the equality point of the titration of hydrochloric acid with Na hydrated oxide was determined to be at a pH of 7. 0. The titration of acetic acid with Na hydrated oxide resulted in the equality point of found pH 8. 6. And the titration of ammonium hydroxide with hydrochloric acid had an equality point at a pH of 5. 8. The methyl ruddy index and the phenolphthalein index were unsuitable because their pH ranges for their colour alterations did non cover the equality points of the tests in which they were used. However. the methyl ruddy index is more suited. since it’s pH scope is closer to the equality points of the titrations.