Introduction Strong acids and strong acids both dissociate completely in water forming ions. However, strong acids donate a proton to form H3O+ along with a conjugate base and strong bases accept a proton to form OH- along with a conjugate acid. The chemical behavior of acids and bases are opposite. When they are together, their ions cancel out and form a neutral solution. In this experiment, HCl and NaOH will react to form NaOH and H2O with these two steps: The overall reaction is: Both Na+ and Cl- ions combine to form NaCl. H3O+ and OH- ions combine to form H2O. H2O is part of the overall reaction because it is always the product in acid-base neutralization reaction. Based on this reaction, by adding a known base, the amount of acid can …show more content…
The mass of vinegar used during the experiment was 4.108 grams. It was determined that there were .003129 moles of CH3COOH in the vinegar sample. Using this information and the molar mass of CH3COOH, which was 60.05 g/mol, the mass of acetic acid in the vinegar was calculated: 4.Vinegar is a 5% aqueous solution of acetic acid. Since the mass of acetic acid within the vinegar was calculated as .18789 g in step 3, the percent of CH3COOH was calculated using the following equation: To calculate the percent error, the experimental value of 4.5% acetic acid in vinegar was subtracted by the theoretical value of 5% and divided by 5% to yield a percent error of 8.54%. The following is a copy of the calculations done using decimals: 5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH. Half of this value is 12.63 mL. By interpolating the graph, the pH at this volume was 4.80, which is equivalent to the pKa of acetic acid. According to the tabulated data, the pKa was 4.90 at 15 mL of NaOH. At this point, the change in pH with respect to volume was minimal since these values were far from the equivalence point, which occurred experimentally at 27.41 mL. This can also be seen on the graph as the plateau before the inflection point occured. To calculate the Ka of the acid, the following formula is …show more content…
The calculated value was 1.6 x 10^-5. Conclusions The resulting Ka of the acetic acid from this experiment’s calculations was consistent with the experimental results. The experimental percent of CH3COOH was calculated at 1.6 x 10^-5, while the actual value was 1.8 x 10^-5. The calculated value is much lower because the pH read from the graph at half of the equivalence was higher than the actual value. To have gotten a 0% error between the experimental and actual value for CH3COOH, the pH would have been measured at about 4.75, which is slightly more acidic than 4.80. The percent error was calculated to determine how accurate the Ka of acetic acid was: Since the calculations yielded a 20% error, this shows that experimental error occurred during the experiment. Factors that could have affected the results included improper reading of the meniscus for volume of NaOH, not allowing the NaOH to fully drip into the buret after removing the funnel, adding too much acetic acid after the indicator flashed pink to get an inaccurate equivalence point, and not allowing the solution in the beaker to mix thoroughly to get an accurate reading from the pH
To find the mass percent of acetic acid in vinegar, the molar mass of acetic acid is 60.05 g/mole, and 1.00 g/mol of density, then 0.96 mol×(60.05 g/1 mol) = 57.65 g 57.65 g/1000= 0.0576 ×100 = 5.76% The average mass % of acetic acid in vinegar = 5.53% The average % of acetic acid was 5.53%, which is close to the acidity of the vinegar that was taken in the lab which was 5%, for the different percent we had it could be because of errors in calculations or errors in collecting data. 3.
At this point I found that if the hot plate was at 147˚C the solution boiled more vigorously, meaning that my ketone hade a boiling point of 147˚C, which was close to the known boiling point value for 3-heptanone , 146˚C. The hot plate was turned down after this was noticed. After the solution was heat, approximately for five minutes, the mass was found for the bottom layer, which was 2.27g, and the percent yield was calculated. The percent yield was determined by taking the mass of the final ketone and dividing it by the original mass of the alcohol.
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.
Sulfuric acid was added to the distilled water, drop by drop until the solution had a pH of about 4.0. The bottle was labeled “Acid Rain” with a sharpie. Container 3: Alka-seltzer 650mL of water was added to the container 6 alka-seltzer tablets were dissolved into the water.
If I had a household product labeled sodium bicarbonate, I would add an acidic substance and expect bubble to be created. As we know acid reacts with bubbles when combined with sodium bicarbonate. 2. Write the chemical equation for the reaction in well A6. B BoldI ItalicsU Underline Bulleted list Numbered list Superscript Subscript3 Words NaOh + AgNO3>>>>NaNO3 + AgOH 3.
3. Upon adding 20 drops of NaOH, a white precipitate was formed signifying acidic impurity. In the second NaOH mixture, about 20 drops were administered and no precipitate formed indicating that the ample is more pure than before. Data: Weight of flask = 75.10 grams Weight of the flask with solids =
In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /
Firstly, because the NaHCO3 compound was not stored in a sealed container, therefore dust particles could have changed the results, and making the product impure. Also, there are uncertainties associated with the instruments used in this experiment. This, if the products were measured slightly more than should be, this could have affected the concentrations of the solutions, and therefore causing a larger
CLAIRE MUNTING 29/01/2018 Criterion C EFFECTS OF SURFACE AREA OF CALCIUM CARBONATE UPON RATE OF REACTION Calcium Carbonate Chips 1 Introduction: Within the current investigation, the effects of the surface area of Calcium Carbonate (CaCO3) in combination with Hydrochloric acid (HCl) upon its rate of reaction. CaCO3, commonly referred to as limestone, is an organic substance and is, in a sense, the crystallised “carbonic salt” of the element, calcium2. In addition to being a salt, the pH level of Calcium Carbonate is 9.91, and it is therefore, a basic substance, due to the fact that it is comprised of a pH level higher than 7, which is neutral3. HCl, however, is the bodily acid found in the stomach of human beings.
The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
The objective of the experiment that the class and myself conducted dealt with the solubility and stress of the beet tonoplast implemented by the solvent chosen. When conducting this experiment, one of the objectives was to learn the different areas of the cell we are dealing within the beet and where to observe the damage done to the beet tissue, if any is done at all. In this case, many of the different areas we were testing and observing were found inside the cellular membrane. The membrane’s function is to separate and organize the myriad of reactions within cells as well as allow communication with the outer-surrounding environment of the membrane (Texas A&M International Univ Biology Lab Manual).
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:
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
Introduction Buffer is a solution that resists a change in pH when bases or acid are added. Solutions that are acidic contain high concentrations of hydrogen ions (H+) and have pH values less than seven. Buffer usually consist of a weak acid, and its conjugate base or a weak base and its conjugate acid. The function of buffer is to resist the changes in hydrogen ion concentration as a result of internal and environmental factor. This buffer experiment is important so that we relies the important of buffer in our life.
It was calculated and found that the concentration of benzoic acid was higher at 30℃ (0.0308M) than at 20℃