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Difference Between Acid And Base Titration

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Chemical Engineering
School of Engineering and Physical Sciences

FIRST YEAR LABORATORY REPORT
COVER SHEET

Name: Farid Mustafayev Experiment Title: Acids and Basis: Determination of Concentration Using Titration Date of Experiment:

Demonstrator: Sevda Zargarova

Supervisor: Ziyafaddin Asadov

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The experiment consisted of four main parts and in each part given acids and bases were different compared to previous one in terms of their strengths and weakness. In each reaction between the different type of acid and base we had unknown variables and to identify one of them the graph which represented dependence of pH of mixture on added substance was plotted. This graph allowed us to find out the volume of added substance in the equilibrium point. the equilibrium point of reaction is the point where the amount of base added was exactly the correct amount to have the moles of base completely react with the moles of acid and pH becomes 7 which represents neutralized environment. Since there is a 1:1 stoichiometry in the chemical reaction, as we know that for every one molecule of base that was needed to react with one molecule of acid. So the moles of bases and acids become same and as a result we can obtain equation which is given below to figure out unknown concentration of acid. C_acid*V_acid=C_base*V_base Finally, the obtained results from experiment compared with the theoretical values to sure from the punctuality of

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993 Words | 4 Pages

Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029

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778 Words | 4 Pages

With the data collected from this chemical reaction percentage yield of the precipitate can be calculated to determine the effects of impurities in the reactants on percentage yield. PRE-LAB QUESTIONS 1. What is the balanced chemical equation for the reaction in this investigation? 2. Which reactant is considered impure in this investigation?

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1270 Words | 6 Pages

The unknown compound was first reacted with an acid. To begin, 0.50 grams of KCl was mixed with 5 mL of water. Then, 1 mL of 6 M H2SO4 was added to the solution. Secondly, the unknown compound was reacted with a base. Exactly 0.50 grams of KCl was mixed with 5 mL of water, and 1 mL of 1 M NaOH was added to the solution next.

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759 Words | 4 Pages

The temperature of the sulphuric acid was not measured throughout the experiment, however the room in which the experiment was conducted was kept constant, so the chance of any large error due to unknown temperature of the sulphuric acid was most likely reduced. The amount of sulphuric acid used was also controlled by measuring 100mL with a 100mL measuring cylinder to ensure that the results would be consistent. The volume of the agar cubes was calculated from the surface area of each agar cube, both before and after they had been in the sulphuric acid. This increased the reliability of the results as it allowed the rate of diffusion of the sulphuric acid into the agar cubes to be calculated more accurately. The concentration of the acid was 0.1M, which was placed in all three agar cubes to maintain consistency of results.

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653 Words | 3 Pages

This was done for all 4 trials and then the mean molarity and standard deviation of acetic acid was calculated to be 0.8438 ± 0.0814. The 0.4842 grams of acetic acid is derived from the 8.063*10-3 moles. Then the percent of acetic acid in vinegar is calculated by dividing 0.4842 g of acetic acid by 10.214 g of vinegar and multiplied by 100% to get a 4.741% of acetic acid in vinegar. The mean molarity and standard deviation of acetic acid was 0.8438 ±

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470 Words | 2 Pages

In the chemical reaction of hydrochloric acid, HCl, and sodium hydroxide, NaOH, the theoretical ratio is one HCL for every one NaOH. By then taking a known measurement of HCl, in this case 10mL, in addition to the phenolphthalein indicator and deionized water, and titrating it with NaOH, the amount of NaOH that causes the reaction to be complete was calculable given that we know when the reaction was complete. After calculating the moles of each substance added to complete the reaction, the experimental ratio was computed. In this case, the experimental ratio of NaOH to HCl is 1.14:1 as shown in Table 2. Overall, the titration enables the ability to find the exact amount needed for a reaction.

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1167 Words | 5 Pages

Controlled Concentration of amylase Amount of amylase/starch Ph of the amylase/starch The concentration of the Amylase was kept at 1% at at times throughout the experiment. 5cm3 of both will be used in each reaction. pH of the Amylase/starch will be kept the same.

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845 Words | 4 Pages

Discussion: End point in this acid-base titration experiment refers to the point where the chemical reaction has reached its conclusion and no additional titrant should be added. The end point of this experiment can be obtained when the indicator used changes colour. For example colourless to light pink when phenolphthalein is used and red to orange and subsequently yellow when methyl orange is used. Equivalence point, also known as stoichiometric point in an acid-base titration refers to the point where the reacting acid and base is in equal proportions. In the graph plotted, pH against volume of NaOH, the graph becomes nearly vertical at the equivalence point.

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1001 Words | 5 Pages

IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75

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800 Words | 4 Pages

INTRODUCTION The concept of chemical equilibrium was developed after Berthollet (1803) found that some chemical reactions are reversible. For any reaction mixture to exist at equilibrium, the rates of the forward and backward (reverse) reactions are equal. In the following chemical equation with arrows pointing both ways to indicate equilibrium, A and B are reactant chemical species, S and T are product species, and α, β, σ, and τ are the stoichiometric coefficients of the respective reactants and products: α A + β B ⇌ σ S + τ T

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1489 Words | 6 Pages

The equation of the reaction between sodium hydroxide and ethanoic acid is as follows: CH3COOH + NaOH → CH3COONa + H2O We can measure the end point of titration process and we can also measure the amount of reactants. The concentration of ethanoic acid in the vinegar can be determined through stoichiometric calculations, Using the values obtained from the titration, and also the chemical equation as a reference. Phenolphthalein indicator is used in this acid-base titration Equipment and materials:

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1346 Words | 6 Pages

Practical I: Acid-base equilibrium & pH of solutions Aims/Objectives: 1. To determine the pH range where the indicator changes colour. 2. To identify the suitable indicators for different titrations. 3.

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1187 Words | 5 Pages

A drop of methyl red was added. Also, a 0.01M hydrochloric acid (HCl) was added in a dropwise manner from a syringe until the color of the solution matches that of the first test tube setup. The volume of the HCl used was recorded for the determination of the ionization constant of

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3578 Words | 15 Pages

Aim: To find out the relationship between the greater concentration of sodium thiosulfate when mixed with hydrochloric acid and the time it takes for the reaction (the time it takes for the solution to turn cloudy) to take place and to show the effect on the rate of reaction when the concentration of one of the reactants change. Introduction: The theory of this experiment is that sodium thiosulfate and hydrochloric acid reach together to produce sulfur as one of its products. Sulfur is a yellow precipitate so, the solution will turn to yellow color while the reaction is occurring and it will continue until it will slowly turn completely opaque. The reaction of the experiment happens with this formula: “Na2 S2 O3 + HCL =

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767 Words | 4 Pages

That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.

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