using eglish to report

1    LAB REPORT ON ACID-BASE TITRATIONS

 Aim:

 To determine the weight-to-volume percent of acetic acid present in commercial vinegar and the weight-to-volume percent of ammonia present in window cleaners. Acid-base titrations can be conducted using the above samples against standardized aqueous NaOH and standardized aqueous HCl solutions respectively.

2.      Introduction:

Acetic acid, commonly known as ethanoic acid CH₃COOH, is found in commercial vinegar. It is a weak organic acid that gives vinegar its characteristic pungent smell, sour taste and slight acidity. Being an acid, vinegar is used as a mild disinfectant in cleaning agents as well as a food preservative. Ammonia, NH_3, is used in household and industrial cleaning agents due to its ability to solubilize grease.  In this experiment, we shall use acid-base titration to determine the weight-to-volume percentage of acetic acid and ammonia. An acid-base titration is a process of obtaining quantitative information of a sample using an acid-base reaction by reacting with a certain volume of reactant whose concentration is known. A suitable indicator for determining the equivalence point is used to indicate the end point of an acid-base titration. To achieve accurate results, multiple titrations will be carried out until duplicate determinations agree to within 0.05mL of each other.

4. Results & Calculations:

4.1 Standardization of NaOH

Molarity of KHC₈H₄O₄ is 0.05044 moldm^-3   = = 0.05044 moldm^-3)

Titration	1	2	3
Final volume of titrant/cm3	5.00	10.00	14.80
Initial volume of titrant/cm3	0.00	5.00	10.00
Volume of titrant used/cm3	5.00	5.00	4.80
		

KHC₈H₄O₄ (aq)   +  NaOH (aq)                      KHC₈H₄O₃ ^- Na⁺  (aq)  +  H₂O (l)

Average volume of NaOH titrate used = ½ (5.00 + 5.00)

     = 5.00 cm³

No. of moles of KHC₈H₄O₄   reacted = 0.01 × 0.05044

     = 5.044 x 10^-4 mol

From the equation, mole ratio of NaOH ≡ KHC₈H₄O₄   is 1:1,

No. of moles of NaOH reacted                 = 5.044 x 10^-4 mol

Concentration of NaOH  = moles of NaOH/ volume of NaOH

                                                    = 5.044 x 10^-4 /

    = 0.1009 moldm^-3   

Molarity of NaOH is: 0.1009moldm^-3

4.2 Standardization of HCl

Titration	1	2	3
Final volume of titrant/cm3	10.80	10.75	10.70
Initial volume of titrant/cm3	0.00	0.00	0.00
Volume of titrant used/cm3	10.80	10.75	10.70
			

HCl (aq)   +  NaOH (aq)                     NaCl (aq)  +  H₂O (l)

Average NaOH titrate used  = ½ (10.70 + 10.75)

     = 10.73 cm³

No. of moles of NaOH   reacted  = (10.73 ÷ 1000) × 0.1009

     = 1.082 x 10^-3 mol

From equation, mole ratio of NaOH ≡ HCl   is 1:1,

no. of moles of HCl reacted                     = 1.082 x 10^-3 mol

Hence concentration of HCl  = 1.082 x 10^-3 (1000 ÷ 10.00)

     = 0.1082 moldm^-3   

Molarity of HCl is: 0.1082 moldm^-3   

4.3 Determination of acetic acid in vinegar

Titration	1	2	3
Final volume of titrant/cm3	12.15	12.20	12.20
Initial volume of titrant/cm3	0.00	0.00	0.00
Volume of titrant used/cm3	12.15	12.20	12.20
			

CH₃COOH (aq)   +  NaOH (aq)                     CH₃COO^-Na⁺ (aq)  +  H₂O (l)

Average volume of NaOH titrate used = ½ (12.20 + 12.20)

     = 12.20 cm³

No. of moles of NaOH reacted              = (12.20  ÷ 1000) × 0.1009    

                = 1.231 x 10^-3 mol

From equation, mole ratio of NaOH ≡ CH₃COOH   is 1:1,

no. of moles of CH₃COOH reacted           = 1.231 x 10^-3 mol

No. of moles of CH₃COOH in volumetric flask (diluted)  = (100 ÷ 10) × 1.231 x 10^-3

       = 1.231 x 10^-2 mol

No. of mole of CH₃COOH in 25mL (undiluted)  = 1.231 x 10^-2 mol

Molarity of acetic acid = (1.231 x 10^-2 mol) ÷(25 x 10^-3)=0.4924 moldm^-3

Molar mass of CH₃COOH = 2(12.01) + 4(1.0079) + 2(16.00)

= 60.0516 g mol^-1

Therefore, weight-volume percentage of CH₃COOH in vinegar  

= (0.4924 x 60.0516 ÷ 1000) × 100% = 2.96%

Percentage of acetic acid in the vinegar is: 2.96%

Indicator used: phenopthalein

4.4 Determination of ammonia in window cleaner

Titration	1	2	3
Final volume of titrant/cm3	4.85	9.65	4.85
Initial volume of titrant/cm3	0.00	4.85	0.00
Volume of titrant used/cm3	4.85	4.80	4.85

NH₃ (aq)   +  HCl (aq)                     NH₄⁺Cl^- (aq)

Average volume of HCl titrate used = ½ (4.85 + 4.85)

     = 4.85 cm³

No. of moles of HCl reacted = 0.1082 x (4.85 x10^-3)

         = 5.248 x 10^-4 mol

From equation, mole ratio of HCl ≡ NH₃   is 1:1,

no. of moles of NH₃ reacted = 5.248 x 10^-4 mol

No. of moles of NH₃ in volumetric flask (diluted) = (100 ÷ 10) × 5.248 x 10^-4 mol

       = 5.248 x 10^-3 mol

No. of mole of NH₃ in 10mL (undiluted)   = 5.248 x 10^-3 mol

Concentration of NH₃   = (5.248 x 10^-3) ÷ (10 x 10^-3) = 5.248 x 10^-1 moldm^-3

Molar mass of NH₃ = 14.01 + 3 (1.0079)

= 17.0337 g mol^-1

Therefore, weight-volume percentage of NH₃ in window cleaner  

= (17.0337 x 5.248 x 10^-1)/ 10³ × 100%

= 0.894%

Percentage of ammonia in the window cleaner is: 0.894%

Indicator used: methyl red

5. Discussion:

The main reaction in the series of acid-base titrations is neutralization. During neutralization, the acid and base reacts to form a salt and water (i.e. H⁺ + OH^-          H₂O). As the all the reactants are colourless, an appropriate indicator must be added to indicate the colour change at the end point.

For Experiment 3.3, the indicator phenolphthalein was chosen as it was a weak acid-strong base titration, as acetic acid is a weak acid and NaOH is a strong base. At equivalence point, the resulting solution contains only CH3COONa. Thus salt hydrolysis occurs and the pH at equivalence point is > 7. Phenolphthalein changes colour at a pH range of 8.2 – 10.0. The (pH)end point falls within the pH for colour change of phenolphthalein in the titration. The colour of phenolphthalein changes explicitly from pink to colourless and the colour change is over a narrow pH range. This further enhances its suitability as an indicator for this titration. Salt hydrolysis that raises equivalence point to > 7 pH : CH3COO-  +  H2O          CH3COOH  +  OH-

* The pH titration curve of weak acid (CH3COOH) and strong base (NaOH)

However, for Experiment 3.4 a different indicator was chosen. Methyl red was used instead as it was a case of a strong acid-weak base titration, as HCl is a strong acid and NH3 is the weak base. At equivalence point, the resulting solution contains only NH4Cl. Thus salt hydrolysis occurs and the pH at equivalence point is < 7. Methyl red changes colour over a pH range of 4.8 – 6.0. The (pH)end point falls within the pH range for colour change of methyl red in this titration. The colour of methyl red changes explicitly from red to yellow and the colour change is over a narrow pH range. This further enhances its suitability as an indicator for this titration. Salt hydrolysis that lowers equivalence point to < 7 pH : NH4Cl  +  H2O           NH3  +  H3O+

* The pH titration curve of weak base (NH3) and strong acid (HCl)

Strong acid-weak base and weak acid-strong base titrations are employed in this experiment instead of weak acid-weak base titrations as the hydrolysis of a salt of a weak acid-weak base is incomplete. The hydrolysis of the conjugate base results in an acidic solution and at the same time, hydrolysis of the conjugate acid of the salt gives rise to a basic solution. Hence, the resulting solution at equivalence point can be neutral, acidic or basic depending on which hydrolysis occurs at a greater extent. This means that calculations have to be done preceding the titration in order to select the appropriate indicators. Therefore, weak acid-weak base titration is not commonly used to determine the quantitative information of an unknown sample.

While titrating, the conical flask has to be swirled constantly throughout to ensure that the contents are mixed evenly. Also, some titrant may drip onto the sides of the conical flask and may not react with the solution in the conical flask. This reduces the accuracy of the results as extra titrant would be used to achieve the equivalence point. To prevent this, deionised water can be used to wash down the unreacted titrant when nearing the equivalence point and the conical flask should be swirled before continuing with the titration. To achieve consistent results, one important factor is to add in the titrant in a drop wise manner before the equivalence point.

Deionised water was used not only for diluting the sample but also for washing the apparatus, instead of using tap water. This is because deionized water is pH neutral and will not negatively affect the results of the acid-base titrations. This is important as acid-base titrations are pH sensitive reactions.

Multiple titrations were carried out until duplicate determinations agree to within 0.05mL of each other. This ensures that the results are not once-off outliers. By averaging the two most consistent and accurate results, the probability of incurring random errors was accounted for and decreased.

Other factors that would affect the accuracy of the results include parallax error when taking the readings and making sure that no air bubbles are present within the pipette and burette while titrating.

6. Conclusion:

Using acid-base chemistry, repeated titrations of vinegar sample and window cleaner sample was carried out against standardized solutions of NaOH and HCl respectively. Through stoichiometric relations, the concentration and weight-to-volume percentage of acetic acid in vinegar and ammonia in window cleaner can be found. The percentage of acetic acid in the vinegar sample studied was deduced to be 3.96%. The percentage of ammonia in the window cleaner sample is 0.894%.

using english to calculate

What do cookies and chemistry have in common? Many things, it turns out! A balanced chemical equation is the recipe for a reaction: it contains a list of all the reactants (the ingredients) and products (the cookies) as well as their relative proportions.

Using a balanced chemical equation to calculate amounts of reactants and products is called stoichiometry. It is a super technical-sounding word that simply means using ratios from the balanced equation. In this article, we will discuss how to use mole ratios to calculate the amount of reactants needed for a reaction.

Balanced reactions and mole ratios

The stoichiometric coefficients are the numbers we use to make sure our equation is balanced. We can make ratios using the stoichiometric coefficients, and the ratios will tell us about the relative proportions of the chemicals in our reaction. You might see this ratio called the mole ratio, the stoichiometric factor, or the stoichiometric ratio. The mole ratio can be used as a conversion factor between different quantities.

Problem solving tip: The first and most important step for all stoichiometry problems is the same no matter what you are solving for—make sure your equation is balanced! If the equation is not balanced, the mole ratios will be wrong, and the answers will not be correct.

For example, the stoichiometric coefficients for the following balanced equation tell us that 1 mole of Fe₂O₃ wiil react with 2 moles of Al to yield 2 moles of Fe and 1 moles of AL₂O₃.

            Fe₂O₃(s) + 2Al (s)               2Fe(l) + Al₂O₃

If we have a known mass of the reactant Fe₂O₃, we can calculate how many moles of Al we need to fully react with the Fe₂O₃ using the ratio of their coefficients:

Mole ratio between Al and Fe₂O₃ = 2mole Al/1mole Fe₂O₃

Example: Using mole ratios to calculate mass of a reactant

For the following unbalanced reaction, how many grams of NaOH will be required to fully react with 3.10 grams of H₂SO₄ ?

NaOH(aq)+H₂SO₄(aq)→H₂O+Na₂SO₄(aq)       Not balanced!

For this reaction, we have 1 Na and 3 H on the reactant side and 2 Na and 2 H on the product side. We can balance our equation by multiplying NaOH by two—so that there are 2 Na on each side—and multiplying H₂O on both sides. That gives the following balanced reaction:

2NaOH(aq)+H2SO4(aq)→2H2O+Na2SO4(aq)    Balanced!

Once we have the balanced equation, we can ask ourselves the following questions:

• For which reactant(s) do we already know the amount of the chemical?
• What are we trying to calculate?

In this example, we know the amount of H₂SO₄ is 3.10 grams, and we would like to calculate the mass of NaOH. Armed with the balanced equation and a clear sense of purpose—hopefully—we can use the following strategy to tackle this stoichiometry problem:

Step 1: Convert known reactant amount to moles.

The known quantity in this problem is the mass of H₂SO₄. We can convert the mass of H₂SO₄ to moles using the molecular weight. Given that the molecular weight of H₂SO₄ is 98.09 g/mol, we can find the moles of H₂SO₄

Step 2: Use mole ratio to find moles of other reactant.

We are interested in calculating the amount of NaOH so we can use the mole ratio between NaOH and H₂SO₄. Based on our balanced chemical equation, we need 2 moles of NaOH for every 1 mole of H₂SO₄, which gives the following ratio:

Mole ratio between NaOH and H₂SO₄=2mol NaOH/1mol H₂SO₄ or 1mol H₂SO₄/2mol NaOH

Each format gives a different answer! However, only one ratio will allow the units of H₂SO₄ to cancel out properly.

Step 3: Convert moles to mass.

We can convert the moles of NaOH from Step 2 to mass in grams using the molecular weight of NaOH:

We will need 2.53 grams of NaOH to fully react with 3.10 grams of H₂SO₄ in this reaction.

Shortcut: We could also combine all three steps into a single calculation, with the caveat that we should pay extra close attention to our units. In order to convert the mass of H₂SO₄ to mass of NaOH, we could solve the following expression:

If we look carefully at the expression, we can break it down into steps 1 to 3 above. The only difference is that instead of doing each conversion separately, we did them all at once.

Summary

The coefficients from the balanced chemical reaction tell us the proportions of the reactants and products. We can use ratios of the coefficients to convert between amounts of reactants and products in our reaction.

chemistry reaction in daily life

Chemistry happens in the world around you, not just in a lab. Matter interacts to form new products through a process called a chemical reaction or chemical change. Every time you cook or clean, it's chemistry in action. Your body lives and grows thanks to chemical reactions. There are reactions when you take medications, light a match, and take a breath. Here's a look at 8 chemical reactions in everyday life. It's only a small sampling since you see and experience hundreds of thousands of reactions each day.

1.Photosynthesis Is a Reaction To Make Food
Plants apply a chemical reaction called photosynthesis to convert carbon dioxide and water into food (glucose) and oxygen. It's one of the most common everyday chemical reactions and also one of the most important since this is how plants produce food for themselves and animals and convert carbon dioxide into oxygen.

2.respiration Is a Reaction With Oxygen
Aerobic Cellular Re Aerobic cellular respiration is the opposite process of photosynthesis in that energy molecules are combined with ​the oxygen we breathe to release energy needed by our cells plus carbon dioxide and water. Energy used by cells is chemical energy in the form of ATP.
Here is the overall equation for aerobic cellular respiration:
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy (36 ATPs)

3.Combustion Is a Type of Chemical Reaction

Combustion is a chemical reaction in everyday life. WIN-Initiative / Getty Images
Every time you strike a match, burn a candle, build a fire, or light a grill, you see the combustion reaction. Combustion combines energetic molecules with oxygen to produce carbon dioxide and water.
For example, the combustion reaction of propane, found in gas grills and some fireplaces, is:
C3H8 + 5O2 → 4H2O + 3CO2 + energy
       

4.Rust Is a Common Chemical Reaction

Over time, iron develops a red, flaky coating called rust. This is an example of an oxidation reaction. Other everyday examples include formation of verdigris on copper and tarnishing of silver.
Here is the chemical equation for the rusting of iron:
Fe + O2 + H2O → Fe2O3. XH2O
           

5.Mixing Chemicals Causes Chemical Reactions

Baking Powder and baking soda perform similar functions during baking, but they react differently with the other ingredients so you can't always substitute one for the other.
If you combine vinegar and baking soda for a chemical volcano or milk with baking powder in a recipe you experience a double displacement or metathesis reaction (plus some others). The ingredients recombine to produce carbon dioxide gas and water. The carbon dioxide forms bubbles in the volcano and helps baked goods rise.
These reactions seem simple in practice but often consist of multiple steps. Here is the overall chemical equation for the reaction between baking soda and vinegar:
HC2H3O2(aq) + NaHCO3(aq)  → NaC2H3O2(aq) + H2O() + CO2(g)                                

6.Batteries Are Examples of Electrochemistry

Batteries use electrochemical or redox reactions to convert chemical energy into electrical energy. Spontaneous redox reactions occur in galvanic cells, while nonspontaneous chemical reactions take place in electrolytic cells.

7.Acid-Base Reactions- Everyday Chemical Reaction

Whenever you combine an acid (e.g., vinegar, lemon juice, sulfuric acid, muriatic acid) with a base (e.g., baking soda, soap, ammonia, acetone), you are performing an acid-base reaction. These reactions neutralize the acid and base to yield salt and water.
Sodium chloride is not the only salt that may be formed. For example, here is the chemical equation for an acid-base reaction that produces potassium chloride, a common table salt substitute:
HCl + KOH → KCl + H2O
Soaps and Detergents - Everyday Chemical Reactions

8.Soaps and detergents clean by way of chemical reactions. Soap emulsifies grime, which means oily stains bind to the soap so they can be lifted away with water. Detergents act as surfactants, lowering the surface tension of water so it can interact with oils, isolate them, and rinse them away.