Introduction:
A precipitation reaction can occur when two ionic compounds react and produce an insoluble solid. A precipitate is the result of this reaction. This experiment demonstrates how different compounds, react with each other; specifically relating to the solubility of the compounds involved. The independent variable, will be the changing of the various chemical solutions that were mixed in order to produce different results. Conversely the dependent variable will be the result of the independent variable, these include the precipitates formed, and the changes that can be observed after the experiment has been conducted. The controlled variable will be the measurement of ten droplets per test tube.
Purpose:
To predict and the test,
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various possible precipitation reactions. Hypothesis: Five out of the ten tests will produce a precipitate and these will be tests 1,2,7,8 and 10. All of which will produce a variety of different insoluble compounds (precipitates), others will not have visibly changed, or at least not significantly. Materials: Solutions of silver nitrate, sodium chloride, sodium hydroxide, barium nitrate and copper sulfate in ‘dropper bottles’ 10 small test-tubes Test-tube rack Gloves Procedure: 1. Construct a table, with each combination of possible tests then predict which of the ten tests will result in precipitates. 2. Place 10 drops of solution A in the test-tube then 10 drops of solution B, in the same tube and record the result, 3. Repeat step two, for the remaining nine tests. Results: Solution A Solution B Prediction of precipitation Before mixing After mixing Name of precipitate Silver Nitrate Sodium Chloride Yes Clear Liquid Powdery white substance, colloid Silver Chloride Silver Nitrate Sodium Hydroxide Yes Clear Liquid Powdery brown substance, brown colloid Silver Hydroxide Silver Nitrate Barium Nitrate No Clear Liquid Clear liquid- no change. N/A Silver nitrate Copper sulfate No Clear Liquid Clear liquid- no change. N/A Sodium Chloride Sodium Hydroxide No Clear Liquid Clear liquid- no change. N/A Sodium Chloride Barium Nitrate No Clear Liquid Clear liquid no change N/A Sodium Chloride Copper Sulfate Yes Clear Liquid Blue green transparent liquid Copper Chloride Barium Nitrate Copper Sulfate Yes Clear Liquid White colloid Barium sulfate Barium Nitrate Sodium Hydroxide No Clear Liquid Cloudy White colloid N/A 9.Copper sulfate Sodium Hydroxide Yes Clear Liquid blue. White cloudy, blue colloid Copper Hydroxide Discussion: Analyse the solubility rules in table 5.2.2 on page 152 to work out what has precipitated from each solution\ Construct word equations and formula equations to describe what is happening in each case where a reaction occurred. 1. Silver Nitrate + Sodium Chloride Silver Nitrate (aq) + Sodium Chloride (aq) → Silver Chloride(s) + Sodium Nitrate (aq) AgNO₃- (aq) + NaCl-(aq) → AgCl-(s) +NaNO₃-(aq) The precipitate formed is Silver Chloride. 2. Silver Nitrate + Sodium Hydroxide Silver Nitrate (aq) + Sodium Hydroxide (aq) → Silver Hydroxide(s) +Sodium Nitrate (aq) AgNO₃-(aq) + NaOH-(aq) → AgOH-(s) + NaNO₃-(aq) The precipitate formed is Silver Hydroxide 3. Silver Nitrate + Barium Nitrate Silver Nitrate (aq) + Barium Nitrate (aq) → Silver Nitrate (aq) + Barium Nitrate (aq) AgNO₃- (aq) + Ba₂+NO₃ (aq) → AgNO₃- (aq) + Ba₂+NO₃ (aq) No precipitate is formed. 4. Silver Nitrate + Copper Sulfate Silver Nitrate (aq) + Copper Sulfate (aq) → Silver sulfate (aq) + Copper Nitrate (aq) AgNO₃- (aq) + Cu+SO₄²- (aq) → AgSO₄²- (aq) + Cu+NO₃- (aq) No precipitate is formed. 5. Sodium Chloride + Sodium Hydroxide Sodium Chloride (aq) + Sodium Hydroxide (aq) → Sodium Hydroxide (aq) + Chloride (aq) NaCl- (aq) + NaOH- (aq) → NaOH (aq) + NaCl- (aq) No precipitate is formed. 6. Sodium Chloride + Barium Nitrate Sodium Chloride (aq) + Barium Nitrate (aq) → Sodium Nitrate (aq) + Barium Chloride (aq) NaCl- (aq) + Ba₂+NO₃ (aq) → NaNO₃ (aq) - + Ba₂+Cl- (aq) No precipitate is formed. 7. Sodium Chloride + Copper sulfate Sodium Chloride (aq) + Copper sulfate (aq) → Sodium sulfate (aq) + Copper Chloride(s) NaCl-(aq) + Cu+SO₄²-(aq) → NaSO₄²-(aq) + Cu+Cl-(s) The precipitate formed is Copper Chloride 8. Barium Nitrate + Copper Sulfate Barium Nitrate (aq) + Copper Sulfate (aq) → Barium Sulfate(s) + Copper Nitrate (aq) Ba₂+NO₃ (aq) + Cu+SO₄²-(aq) → Ba₂+SO₄²- (s) + Cu+NO₃ (aq) The precipitate formed is Barium sulfate 9. Barium Nitrate + Sodium Hydroxide Barium Nitrate (aq) + Sodium Hydroxide (aq) → Barium Hydroxide (aq) + Sodium Nitrate (aq) Ba₂+NO₃ (aq) + NaOH- (aq) → Ba₂+OH- (aq) + NaNO₃-(aq) No precipitate is formed. 10. Copper Sulfate + Sodium Hydroxide Copper Sulfate (aq) + Sodium Hydroxide (aq) → Copper Hydroxide(s) +Sodium Sulfate (aq) Cu+SO₄² (aq) + NaOH-(aq) → Cu+OH-(s) + NaSO₄²-(aq) The precipitate formed is Copper hydroxide. To understand how a precipitate is formed one must first understand the difference between cations and anions, and their role in precipitation reactions.
Cations are positively charged ions, which are attracted to their negatively charged counterparts, anions. Precipitates can form when these cations and anions combine in aqueous solutions; however, precipitates only form if one of the products of the chemical reaction is not soluble in that solution. Solubility is instrumental in understanding how precipitation reactions occur. This is because solubility rules, determine whether a precipitate can form. A precipitate can form if the cation in the compound is soluble when combined with an anion. For example when the solutions silver nitrate and sodium chloride (reactants) are mixed, silver chloride and sodium nitrate (products) are formed. Following the solubility laws, silver nitrate is the precipitate, as it isn’t …show more content…
soluble. This experiment demonstrated the links between the solubility of the different compounds used (independent variable) and the resulting precipitate. If both solutions were soluble there would be no resulting solid. If one of the solutions was not soluble or had low solubility there would only a resulting precipitate. This result was the dependent variable, however by definition the precipitate was solid. In all cases, with exception to test one, a more accurate description of the precipitate would be not a liquid, as some of the precipitates looked cloudy and coagulated, rather than solidified substance located at the bottom. Another issue which could be improved by for future experiments is the rough estimation of 10 drops per solution (which is the controlled variable), it would be ideal to have an exact measurement, as it would marginalise the differences between test, and try to eliminate as many unnecessary variables as possible. The procedural element of the experiment was very simple, and easy to understand. It did, however, require preparation beforehand, this being the results table; this wasn’t overly complicated however it was time consuming. The task before required predictions to be made in regard to whether each test would yield a precipitate.
This can be done by first finding the products of the chemical reactions, which are found by swapping the anions on each reactant. Once this is done, predictions can be made. The table above, describes the solubility rules, these are used to decide whether a compound will be soluble, and then consequently to this reveal a precipitate. Barium sulfate for example is insoluble and if it was to be mixed with an aqueous compound, barium sulfate would be the precipitate. This is an example of how a prediction can be made, without physically viewing the experiment or given the results. It is also a way of identifying what the precipitate is once the experiment has been
conducted. Conclusion: This experiment, examined various precipitation reactions and attempted to demonstrate how predictions can be made prior to the experiment being conducted. The results proved the hypothesis, which was completely correct, in that the tests 1,2,7,8 and 10 would reveal a precipitate. The hypothesis was also correct in relation to the other tests showing minimal physical changes. The variables involved helped produce the results of the experiment. The independent variable, was the use of various different compounds, which produced the dependent variable (the result, the precipitate or lack thereof). The controlled variables had a minimal impact on the experiment. In summation the experiment fulfilled its purpose and allowed numerous precipitation reactions to be predicted, witnessed and tested.
The purpose for this experiment was to determine why it was not possible to obtain a high percent yield when Calcium Nitrate Ca(〖NO_3)〗_2 with a concentration of 0.101 M was mixed with Potassium Iodate KIO_3 with concentration of 0.100 M at varying volumes yielding Calcium Iodate precipitate and Potassium Nitrate. Filtration was used to filter the precipitates of the solutions. The percent yield for solution 1 was 87.7%, and the percent yield for solution 2 was 70.8%. It was not possible to obtain a high percent yield because Calcium Iodate is not completely soluble and some of the precipitates might have been rinsed back to the filtrates when ethanol was used to remove water molecules in the precipitate.
For this experiment we have to use physical methods to separate the reaction mixture from the liquid. The physical methods that were used are filtration and evaporation. Filtration is the separation of a solid from a liquid by passing the liquid through a porous material, such as filter paper. Evaporation is when you place the residue and the damp filter paper into a drying oven to draw moisture from it by heating it and leaving only the dry solid portion behind (Lab Guide pg. 33.).
During week 1 of this experiment, we recorded common components of fertilizers and then went on to find the chemical formulas involved in creating them. The second week we began the process of comparing three authentic ions we had established in the first week to ion samples to discover other properties they might contain. We decided to discover these different ingredients by preforming a serious of tests, which included placing 0.2g solid of both the authentic and the sample fertilizer separately, in order to establish a constant, and dissolved the fertilizer in 20 mL of water, then checked to see if Mg was present in the sample solution. By setting up a constant and preforming a methodical experiment all on the samples given, we demonstrated the ability to correctly establish and preform an experiment and solve the problem at hand, which was distinguishing the contents of the authentic
In our experiment we utilized the hydrate cobaltous chloride. Hydrates are crystalline compounds in which one or more molecules of water are combined with each unit of a salt. Cobalt (II) chloride hexahydrate is an inorganic compound which is a deep rose color in its hydrated form. As an inducer of
Thorough analysis of the graph displayed enough evidence suggesting that an increase in substrate concentration will increase the height of bubbles until it reaches the optimum amount of substrate concentration, resulting in a plateau in the graphs (figure 2). Hence; supported the hypothesis.
For this experiment, you will add the measured amount of the first sample to the measured amount of the second sample into its respectively labeled test tube then observe if a reaction occurs. In your Data Table, record the samples added to each test tube, describe the reaction observed, if any, and whether or not a chemical reaction took place.
the ones that contains ppt in half, then add 6M NH3 to one set of them
We attached a gas syringe via a plastic tube to a test tube and used a
The last way is to note how long it takes for a precipitate to form
Solubility is defined as the maximum amount of a substance that will dissolve in a given amount of another substance at constant temperature and pressure. Solubility is typically expressed in terms of maximum volume or mass of the solute that dissolve in a given volume or mass of a solvent. Traditionally the equilibrium solubility at a given pH and temperature is determined by the shake flask method. According to this method the compound is added in surplus to a certain medium and shaken at a predetermined time. The saturation is confirmed by observation of the presence of un-dissolved material. Saturation can also be reached if the solvent and excess solute is heated and then allowed to cool to the given temperature. After filtration of the
The purpose of this experiment was to prepare two solutions and use them to perform a precipitate reaction. Then using the results and mass gathered from the experiment, to determine the limiting reagents and to calculate percent yield.
Chemical kinetics is a branch of chemistry that involves reaction rates and the steps that follow in. It tells you how fast a reaction can happen and the steps it takes to make complete the reaction (2). An application of chemical kinetics in everyday life is the mechanics of popcorn. The rate it pops depends on how much water is in a kernel. The more water it has the quicker the steam heats up and causes a reaction- the popping of the kernel (3). Catalysts, temperature, and concentration can cause variations in kinetics (4).
Since Barium (Ba2+) is the only cation in this experiment which has this property, it was determined that the cation for this solution must be Ba2+. The solution was determined to be neutral, based on the litmus test results in Table 2. The anions which form a precipitate with Ba2+ were identified to be Cl-, NO3-, and SO42- using Table 4. Since precipitate was observed to be formed between solution B and solutions C,D, and F in the experiment, the possible anions were compared to determine which one followed these properties. Using the information in table 4, Cl- was declared the anion evident in solution B. Therefore, the solution B must be
borate) and 1.0 g. of sodium hydroxide in 20 mL of warm water. It may
Determination of thermodynamic values allows for analysis of what makes a reaction spontaneous. In this experiment, the equilibrium constant of the crystallization of potassium nitrate as it ionized in water was found and used to determine enthalpy, entropy, and Gibb’s Free Energy of a reaction. The variables were found by by graphing the solubility of potassium nitrate as a function of time and by utilizing relationships based on the van’t Hoff equation. Based on the determined Ksp of 43.4 the average Gibb’s Free Energy over on six trials was -8.4834 kJ/mol with a 510 % error. Relations based on the graph of ln(k) vs. 1/T(K) showed the enthalpy of the reaction to be +34.78 kJ/mol yielding a 2.30% error, and showed the entropy to be +137.4