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Best method for determining silver ions in solution

Extracts from this document...

Introduction

Chemistry Coursework Aim I am going to investigate which is the best method for determining silver ions in solution. I am interested to see whatever a electrode potential method or a chemical method (e.g. titration) is more appropriate, especially at low concentration. Background Information Redox If we consider the following reaction: 2Na(s) + Cl2(g) --> 2Na-Cl+(s) Salt (Sodium Chloride) is formed when Sodium is added to Chlorine gas. Sodium Chloride is an ionic compound, which is made up of Na+ and Cl- ions. The difference between Na(s) and Na+(s) is that an electron is removed from the Na atom and this a positive charge formed as the number of protons, which are positively charge is not equal to the negatively charged electron. But NaCl(s) is not charged overall, so where is the electron gone? If we look at the equation more closely, we can split the reaction into two parts, and this is called half equations: 2Na(s) --> 2Na+(s) + 2e- Cl2(g) + 2e- --> 2Cl-(s) The first equation shows that each sodium atom lost one electron and formed a Sodium ion and an electron. While each chlorine atom gained an electron and form Chloride ions. These two equation is balance and overall there is no electron lose as once a sodium has given up its electron, a chlorine atom will form an ion with this electron. We called this type of reaction Redox - Reduction and Oxidation. The sodium in this example is being oxidized as the atom loses an electron while chlorine is being reduced as it gains one electron. Electrochemical cell and electrode potential In the last example, we saw sodium giving up electron and chlorine accepting electron. But sodium could well be accepting electron while chlorine giving up electron. So there must be something deciding the direction of the electron flows. We can investigate this using half cells. If we combine two half cell together, we can form a electrochemical cells. ...read more.

Middle

Swirl and turn the flask for a minute. Chemical Used Molar Mass (g) Mass required in 1dm3 solution of Concentration 1 mol dm-3 (g) Mass required in 0.25dm3 solution of concentration 1 mol dm-3 (g) NaCl 58.5 58.5 14.625 CuSO4.5H2O 160 + 90 = 250 250 62.5 KNO3 (Saturated) 101.1 101.1 25.3 As I am going to investigate the different concentration of the Ag+ ions in solution. I will use the following instructions which enable me to produce a range of solutions from 0.1M to very dilute: 1. Using the standard 0.1M AgNO3 solution and a pipette, I will pipette 25ml of them into the standard flask (250ml) and then fill the rest up to the line with distilled water. 1. For 0.01M, I will pipette 1/10th of the 1M solution (25ml) and then fill the rest up to the line with distilled water. 1. Repeat the above steps, but take the solution from step 2 until all the required solution is made. By doing this, I am dividing the concentration of the solutions by 1/10th every time. Thus, I will obtain a series of solution of AgNO3(aq) with different concentration. I will also need a series of Sodium Chloride solution with different concentration for titration as well. I will use the above steps. Working out the E(cell) from Nernst Equation: Using a spreadsheet on computer (next page), I am able to work out a set of prediction on what E(cell) value is going to be like using the Nernst Equation at 298K (room temperature) using the following half cells, which all of the E(cell) going to be measured against: 2Ag+ | 2Ag || Cu2+ (1M)| Cu The prediction assume that the concentration of both half cell solution doesn't changes much after I switch on the apparatus as I only used up a very small amount of metal ions and metal. Finding out the E(cell) ...read more.

Conclusion

According to the Nernst Equation, the lines should be parallel with same gradient for each line. The graph suggested that there are some significant change in the E(cell) value when the temperature change. But the change was not as predicted - the higher temperature should decrease the E(cell) value. This may due to the fact that the internal resistance changes inside the solution which affect the current flowing in the system. The following is the results of doing the titration of silver nitrate at different temperature. This set of experiment is done using 0.1M of the silver nitrate titrated against 10ml of 0.1M of sodium chloride solution. The reaction vessel (conical flask) is left in the water bath and get to the temperature which to be investigated. The results shows a more accurate than the original experiment. (We expected that there should be 10cm3 of the sliver nitrate used in the titration). I expected that the results should be more error introduced - the solubility product of the sliver nitrate increases with temperature thus the red precipitation is not appearing unless with more sliver nitrate added to the solution. Conclusion There are different situation where each of this method whould be used in the determination of the silver. The E(cell) method give you an good approximation of the silver ions concentration. But it depends on the concentration and purity of the solution, as impurity does affect electropotential. The E(cell) experiment can be used in all temperature as the E(cell) value can be calculate using Nernst Equation. Chemical method, like titration, is used when there are only limited amount of test solution available. The indicator is very important in the success of the determination as we saw that the indicator can introduce an blank error into the experiment. The temperature of which the determination happening also affects the result. Generally, all the reaction should take place in room temperature as we have found out that at higher temperature, there is a different value obtained. ...read more.

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