Chemistry Lab Report Reports Example

Type of paper: Report

Topic: Acid, Solution, Fat, Reaction, Equation, Sodium, Calcium Carbonate, Calcium

Pages: 9

Words: 2475

Published: 2021/02/20

Back titration of Calcium Carbonate with Sodium Hydroxide.

Introduction
Here the amount of Calcium Carbonate is asked and because it is an insoluble salt "back titration" is used. Titration is a method which it's another name is volumetric method, in this method a reactant is used to bring a reaction to endpoint. In this method, one sample (unknown) value is determined by a standard (known) value.
But "back titration" is needed because here Calcium Carbonate is an insoluble salt and it's not possible to calculate its amount by direct titration.
In this experiment first calcium carbonate was reacted with an excess known hydrochloric acid and the unreacted acid was then reacted with sodium hydroxide and the endpoint of the reaction was determined by a color change, which was indicated by the presence of methyl orange indicator.

The equation for the reaction of this titration was as follows:

CaCO3 + 2HCl → CaCl2 + H2O + CO2
The recorded titration volumes of sodium hydroxide were then used to calculate the moles of calcium carbonate. First of all the volumes of known solution whose concentration is known were used to calculate of moles of excess acid which reacts with sodium hydroxide 1:1, the amount of excess acid was subtracted from initial amount of acid and the consumed value of acid in the reaction with calcium carbonate was calculated , the moles of calcium carbonate will be obtained by dividing moles of consumed acid in two because 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid.

Method

First of all weight exactly 1.6 gr of calcium carbonate sample into a 400 cm3 beaker. Cover the powder with about 20cm3 of deionized water and then carefully pipette in exactly 50 cm3 of the given standard 1 Molar hydrochloric acid solution effervescence is complete, (if the solid does not dissolve, warm gently) transfer the reaction mixture with washings to a 250cm3 volumetric flask. Dilute to the calibration mark with water, then mix well. Pipette out 25cm3 aliquots of this diluted reaction mixture into 250cm3 conical flasks and titrate with the given standard 0.1 Molar sodium hydroxide solution using methyl orange indicator. Resume to adding sodium hydroxide until the color of acid solution changes to yellow. (methyl orange : red → yellow)

Result/Calculations

The average volume of NaOH required to reach the endpoint of reaction was 20.2 cm3
The moles of NaOH reacted with excess HCl : 0.0202×0.1=0.00202 moles

Hydrochloric acid and sodium hydroxide react 1:1; therefore, there were 0.00285moles of HCl in the conical

We transferred 25cm3 from the 250cm3 volumetric. This is 10%. Therefore the excess acid in the 250 = 0.0202 moles of excess HCl
Moles Acid Added initially: n = V x M: 0.05 (L) x 1(M) = = 0.05 moles of HCl
Moles of HCl reacted with the calcium carbonate: 0.05 – 0.0202 = 0.0298moles
Moles = m (g) / Mr → 0.0149 =? / 100.09 → 0.0149 x 100.09 = mass (g) → Mass of Calcium Carbonate =1.491 g
We weighed 1.6g, but the reaction calculated 1.491g
Purity of the calcium carbonate = 1.491/1.6 * 100 = 93.1%

Discussion

The aim of the experiment was to determine the purity of the calcium carbonate by back titration with a standard solution of sodium hydroxide. The average volume of sodium hydroxide required to react with 25cm3 of Hydrochloric acid solution was 28.5 cm3 of sodium hydroxide. The measured titration volumes are reported in Table 1 and were quite reproducible. A methyl orange indicator was used and the endpoint was determined with a color change from red to yellow. From the measured amounts the percentage of calcium carbonate in the given sample t was determined to be 93.1%.

Conclusion

The percentage of calcium carbonate in the given sample t was determined to be 93.1%.

Neutralization Titrations

Introduction
There are two types of acids and bases, strong and weak, strong acids in an aqueous solution separate 100% also strong bases are similar. Strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO4 and HClO3 other usual acids are weak acids. Weak acids in an aqueous solution separate less than 100%. For example, Acetic acid is a weak acid. In the other side, all of the metal hydroxide are strong bases.

Titration curves for Strong acid and base

In the titration of strong acid and strong base, there are three stages and there are needed three types of calculations: 1- pre equivalence 2-equivalence 3-postequivalence. In the pre equivalence stage we calculate the concentration of acid from its initial concentration, in the equivalence stage the hydronium and hydroxide concentration are equal and they are gained directly from separation of water, in the post equivalence stage the concentration of hydroxide is computed from direct analyze of base.

Titration curves for weak acids

"For weak acid there are four distinctly types of calculations to drive a titration curve for a weak acid:
At the beginning the solution contains only the solute acid or base, and the PH is calculated from the concentration of that solute and its dissociation constant.
After various increments of titrant have been added ( in quantities up to , but not including , an equivalent amount), the solution consists of series of buffers. The PH of each can be calculated from the analytical concentrations of conjugate base or acid and the residual concentrations of the weak acid or base.
At the equivalence point, the solution contains only the conjugate of the weak acid or base being titrated (that is, a salt), and the pH is calculated from the concentration of this product.
Beyond the equivalence point, the excess of strong acid or base titrant represses the acidic or basic character of the reaction product to such an extent that the pH is governed largely by the concentration of the excess titrant."(Skoog,West and Holler , Analytical Cemistry 1996)

Henderson-Hasselbalch Equation

A solution containing a weak acid, HA:
HA + H2 ↔ H3O+ + A- Ka= [H3O+][A-]/[HA] (1)
The Henderson-Hasselbalch Equation is an alternative form of equation (1) that is frequently encountered in biological and biochemical literature. It is obtained by expressing each term in the equation in the form of its negative logarithm and inverting the concentration ratio to keep all signs positive.
-log [H3O+]= - log Ka – log cHA/cNaA ; cHA= [HA] , cNaA= [A-]

Therefore,

pH =pKa + log cNaA/cHA (2)
Method
Take 25cm3 of 0.1M hydrochloric acid solution and run in 0.1M sodium hydroxide solution 1 cm3 at a time and note the pH after each addition. When close to the neutralisation point, add the base in smaller aliquots. Plot pH against volume of base added and comment on the shape of the graph obtained.

Equation:

HCl + NaOH → NaCl + H2O
Repeat the experiment using 25cm3 of 0.1M ethanoic (acetic) acid and neutralise with the same sodium hydroxide solution. Plot pH against volume of base added and comment on the shape of the graph obtained.

Equation:

CH3COOH + NaOH → CH3COONa + H2O
Repeat the titration a third time using 25cm3 of 0.05M sodium carbonate solution and titrate against 0.1 M hydrochloric acid. Plot pH against volume of acid added and comment on the shape of the graph obtained.

Equation:

Na2CO3 + 2 HCl → 2 NaCl + H2CO3

Result/Calculation

Result1 : Strong acid - strong base
Result 2: Weak acid - strong base
Result 3: Carbonate bicarbonate system
In the figure from result 1 for a large range changes of NaOH volume until near to 20 ml there is a steady upward trend but the equivalence point in the titration is characterized by major changes in the relative concentration of reagent and analyte the figure shows the equivalence point area with sharp increase in the values of pH, after equivalence area again there is gradual rise in the chart and it begins to level out.

In result2 because weak acid was used it is possible to use from this equation:

Ka = [H3O+]2/cHA → Here PH of acid which there was initially is used, the initial PH = 3.7 and cHA = 0.1 M → Ka= 3.98×10-7
→ pKa= 6.4
At the point of half of neutralization the equation reaches equilibrium , there for concentrations of salt and acid become equal and in the equation of Hesselbalch the logarithmic term become zero and it concludes pH=pKa
In the figure from result 3 for a large range there is steady downward trend but near the equivalence point there is a sharp decrease in the values of pH then again there is gradual decline and it begins to level out.

The equation of ionization in the third part:

CO32− + 2 H2O  HCO3− + H2O + OH−  H2CO3 +2 OH−
Discussion
The aim of these experiments was to get the values of pH in the neutralization titrations and plot pH against volume of base and acid which was used , for these we did three separated experiments , first the stong acid and strong base was used where base was added slowly to the acid and the figure showed a sharp rise in pH near to the equilibrium point, then weak acid was reacted with a strong base where again base was added gradually to the weak acid where and also there was a sharp rise around the equilibrium point and the pKa of weak acid was calculated 6.4, finally sodium carbonate was titrated HCl and the figure of pH shows an steep decline until equilibrium area and around equilibrium point there is a sharp decrease in pH .
Conclusion
The figures of neutralization titrations have a sharp change around equilibrium point and the pKa=6.4 for weak acid and strong base there for the equilibrium point is not always exactly in pH =7.

The Iodine Number of an Oil or Fat

Introduction
Fats and oils are a mixture of triglycerids. Triglycerides are made up of three fatty acids linked to glycerol by fatty acyl esters. Fatty acids are long chain hydrocarbons with carboxyl groups (COOH groups). These fatty acids can be classified into saturated or unsaturated based on the number of double bonds present in the fatty acid. Saturated fatty acids contain only single bond between the carbon atoms and are tend to be solids at room temperature. Unsaturated fatty acids contain double bonds between the carbons atoms in addition to the single bonds present in the fatty acid chain.
Halogens add across the double bonds of unsaturated fatty acids forming an addition compound. Wij's solution (iodine monochloride) is allowed to react with the fat in the dark. The amount of iodine consumed is determined by titrating the iodine released, after addition of excess potassium iodide with standard sodium thiosulphate solution and comparing with a blank in which the fat is omitted.
Fatty acids react with a halogen [iodine] resulting in the addition of the halogen at the C=C double bond site. In this reaction, iodine monochloride reacts with the unsaturated bonds to produce a di-halogenated single bond, of which one carbon has bound an atom of iodine.
After the reaction is complete, the amount of iodine that has reacted is determined by adding a solution of potassium iodide to the reaction product.
ICl + KI ---------------> KCl + I2
This causes the remaining unreacted ICl to form molecular iodine. The liberated I2is then titrated with a standard solution of sodium thiosulfate.
I2+ 2Na2S2O3 -----------------> 2 NaI + Na2S2O4
The reaction between a simple halogen such as iodine and the double bond is rather slow at room temperature, so the use of iodine monochloride "speeds up" the process. The reaction mixture is kept in the dark and the titration carried out as quickly as possible since oxidation of the halogens in light can occur.

The Iodine Number is defined as the number of grams of iodine taken up by 100g of fat.

Method
Weigh into a clean dry conical flask about 0.3g of the sample. Add 10 cm3 of chloroform and then add, using a burette 25cm3 of Wij's solution. Keep the flask in the dark for 30 - 45 minutes for the addition reaction to occur. Shake the solution, add a spatula of potassium iodide and mix well. Add 100cm3 of de-ionized water and titrate to a pale straw color with the 0.1mol
dm-3Solution of sodium thiosulphate provided. Add starch indicator and continue titrating until the blue colour just disappears. Carry out a blank determination. Compare the iodine values of all the samples available and discuss the results in terms of the structures of the materials.
I2No = (B-T) x 0.1 x 127 x 100
1000 x W

Result

The values shows the double bond between carbon atoms in butter is the most
then double bonds in lard is more than oil ,therefor the unsaturated bonds in butter is more than the others and second one is lard and saturated bonds in oil is less than the others.
Discussion
Fats and oils are called triglycerides (or triacylcylgerols) because they are esters composed of three fatty acid units joined to glycerol, a trihydroxy alcohol. Some of fatty acids are unsaturated, if they have unsaturated bonds between carbon atoms they will be unsaturated. The method used to determine the unsaturated bonds is iodine number, which is defined as the number of grams of iodine taken up by 100g of fat. The higher the iodine number, the more C=C bonds are present in the fat.
Conclusion
The result shows the most iodine number for butter, therefor unsaturation amount of butter is most and it is least for oil.

Answer the questions

Define what is meant by the terms ‘Weak Acid’ and ‘Strong Acid’.
According to the Bronsted-Lowry theory, an acid is a proton donor the strong acid have intensive tendency for giving proton.
According to Lewis theory acid is a material dissociated in water and produce H3O+ ion, therefor the strong acid will be separated 100% in the water.

Henderson-Hasselbach normally applies to the weak acid because weak acids

Do not dissociate completely then there is a balance relationship between acid and its salt, therefor it is possible to write equilibrium constant and drive Henderson-Hasselbach equation.

Explain the relationships outlined in the Henderson- Hasselbach Equation.

The Henderson-Hasselbach equation is obtained from equilibrium constant equation for balance reaction of weak acid and its salt:
HA + H2O → H3O+ + A-
Ka = [H3O+]×[ A-]/[HA] → -pKa +pH= log [A-]/[HA]
→ pH = pKa+ log[A-]/[HA]
Identify three differences you would expect to find between the pH profiles graphs of a Strong Acid vs Strong Base reaction and a Weak Acid vs Strong Base reaction.
In titration of strong acid and strong base, the equivalence point there is in the pH of seven, but in titration of weak acid strong base the equivalence point is in the pH more than of seven.
In the weak acid- strong base all of the values of pH in the acidic area are more the values in acidic area of strong acid-strong base reaction.

In the weak acid and stong base the height of the equilibrium area is shorter than another one.

These differences’ reason are because of low dissociation constant which makes the low amount of H3O+ and it will cause higher values of pH. And for realize of equivalence point have to use indicators that changes their color in high pH.

Describe the two main components of a buffer solution that was made by you in one of your practical.

Acid and salt
Explain how the Henderson-Hasselbach equation was used when making up the buffer solution.
Generally for an effective buffer, ratio of molecular concentration and ionic concentration have to be between 0.1 and 10 this rang determine the range of pH.
pH= pKa + log 10-1 = pKa -1
pH= pKa + log 10 = pKa +1
For supply of an effective buffer have to use pH between pKa -1 and pKa +1.

Explain how the Henderson-Hasselbach equation was used to determine the pKin of the indicator in question.

If we use HIn for indicator molecule, we have:
KIn = [H+][In-]/[HIn] → log[KIn] + log [HIn] = log[H+] + log[In-]
→ pKIn = pH- log[In-]/[HIn]

Bibliography:

1. Charles E. Mortimer, 1986. Chemistry. 6 Sub Edition. Wadsworth Pub Co.
2. Bernard Miller, 1980. Organic Chemistry, the Basis of Life. Edition. Addison-Wesley. Miller,B.(1980) Organic Chemistry, The Benjamin/ Cummings Publishing Company, California"
3. Douglas A. Skoog, 1995. Fundamentals of Analytical Chemistry (Saunders Golden Sunburst Series). 7th Edition. Saunders College Pub.
4. Svehla, C. L. Wilson, and D.W. Wilson, Eds.,(1992)Comprehensive Analytical Chemistry.

Cite this page
Choose cite format:
  • APA
  • MLA
  • Harvard
  • Vancouver
  • Chicago
  • ASA
  • IEEE
  • AMA
WePapers. (2021, February, 20) Chemistry Lab Report Reports Example. Retrieved April 14, 2021, from https://www.wepapers.com/samples/chemistry-lab-report-reports-example/
"Chemistry Lab Report Reports Example." WePapers, 20 Feb. 2021, https://www.wepapers.com/samples/chemistry-lab-report-reports-example/. Accessed 14 April 2021.
WePapers. 2021. Chemistry Lab Report Reports Example., viewed April 14 2021, <https://www.wepapers.com/samples/chemistry-lab-report-reports-example/>
WePapers. Chemistry Lab Report Reports Example. [Internet]. February 2021. [Accessed April 14, 2021]. Available from: https://www.wepapers.com/samples/chemistry-lab-report-reports-example/
"Chemistry Lab Report Reports Example." WePapers, Feb 20, 2021. Accessed April 14, 2021. https://www.wepapers.com/samples/chemistry-lab-report-reports-example/
WePapers. 2021. "Chemistry Lab Report Reports Example." Free Essay Examples - WePapers.com. Retrieved April 14, 2021. (https://www.wepapers.com/samples/chemistry-lab-report-reports-example/).
"Chemistry Lab Report Reports Example," Free Essay Examples - WePapers.com, 20-Feb-2021. [Online]. Available: https://www.wepapers.com/samples/chemistry-lab-report-reports-example/. [Accessed: 14-Apr-2021].
Chemistry Lab Report Reports Example. Free Essay Examples - WePapers.com. https://www.wepapers.com/samples/chemistry-lab-report-reports-example/. Published Feb 20, 2021. Accessed April 14, 2021.
Copy

Share with friends using:

Please remember that this paper is open-access and other students can use it too.

If you need an original paper created exclusively for you, hire one of our brilliant writers!

GET UNIQUE PAPER
Contact us
Chat now