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Experiment 29 Molar Solubility (Part A) and Common Ion Effect 实验29摩尔溶解度(一)和共同离子效应

2017-12-26 5页 doc 20KB 33阅读

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Experiment 29 Molar Solubility (Part A) and Common Ion Effect 实验29摩尔溶解度(一)和共同离子效应Experiment 29 Molar Solubility (Part A) and Common Ion Effect 实验29摩尔溶解度(一)和共同离子效应 Experiment 22: Molar Solubility (Part A) and Common Ion Effect (Part B) Background Information A saturated solution of a salt with limited solubility has a reversible reaction and a...
Experiment 29 Molar Solubility (Part A) and Common Ion Effect 实验29摩尔溶解度(一)和共同离子效应
Experiment 29 Molar Solubility (Part A) and Common Ion Effect 实验29摩尔溶解度(一)和共同离子效应 Experiment 22: Molar Solubility (Part A) and Common Ion Effect (Part B) Background Information A saturated solution of a salt with limited solubility has a reversible reaction and an equilibrium expression. The equilibrium constant (K) is called the solubility product. sp ++22-2-AgSO ? 2Ag + SO leads to K = [Ag][SO] 244sp42+-2+-2and Ca(OH) ? Ca + 2OH leads to K = [Ca] [OH] 2sp The aqueous concentration of an aqueous compound (not its individual ions) is called the molar solubility. If there is only one of an ion in the compound’s formula unit, the ion’s concentration is the same as the molar solubility. Otherwise, the ion’s concentration is a multiple of the molar solubility. That is, we can make equations for molar solubilities. +2+-2-[AgSO] = [SO] = (1/2) [ Ag] and [Ca(OH)] = [Ca] = (1/2)[OH] 24(aq)42(aq) -+[OH] is easy to determine with a titration against [HO], and is twice the [Ca(OH)] 32(aq) molar solubility, as seen in the equation above. Substituting the same molar solubility 2+value for [Ca] can provide the solubility product. 2+-2--2-3K = [Ca] [OH] = (1/2) [OH][OH] = (1/2)[OH] sp 2+If [Ca] is increased by adding CaCl, then LeChatelier’s principle tells us that the 2 equilibrium will shift to the left. This equilibrium shift is called the common ion effect. 2+-2+Now [Ca] is no longer the same as the molsol, [Ca(OH)] or (1/2)[OH]. Also, [Ca] 2(aq) can be large enough that the shift is negligible in comparison to the amount from the 2+added CaCl. Therefore, we would not calculate K without knowing [Ca], the 2sp concentration of the added common ion. However, molar solubility can still be -determined with a titration because it can still be calculated as [Ca(OH)] = (1/2)[OH]. 2(aq) As a result, we can experimentally determine the effect of the common ion on the molar solubility, as predicted by LeChatelier’s principle. (Also, see the Common Ion Effect section in the Ch 16 notes.) Procedure Notes 1. Skip steps 1 and 2 of part A, and step 1 of part B. Using beakers, obtain 100 ml of each solution from the supply bench. This quantity will be used to pipet 25-ml samples for each set of three trials. 2. Review technique 16 (pages 29-33) for instruction on pipetting and titrating. 3. Skip the standard deviations on your report sheet. Pre-Lab Hints: 1. a) Molar solubility is the saturated concentration of PbI. 2- Use stoichiometry to the ratio of PbI to I. 2+2-b) Put concentrations of Pb and I into the K expression. sp c) Refer to definition of common ion effect (pp 257-8). Also, you may note that the calculated K for this problem does not match sp the value of K in your textbook (Table 17.1, pg 703). sp 2. Consider how remaining DI water will affect the solution. 3. See footnote on page 259. 4. See step 2 of procedure 16A at the top of page 30. 5. Refer to discussion of molar solubility and common ion effect on pages 257-8. +-6. (moles HO added) = (moles OH in solution) 3--[OH] = (moles OH) / (sample volume) Review molar solubility and K equations for [Ca(OH)]. sp2(aq) 7. Compare endpoint pH (which will be <7) with indicator color change pH’s. You may also want to look in Ebbing and Gammon at Figures 15.10 and 15.11 (pg 642), as well as Fig 16.14 (pg 686). Lab Questions 1. Generally, some solid Ca(OH) is present in the sample. Given that acid removes 2-1OH from the right side of the dissociation reaction, what will happen to the solid and its equilibrium when the acid is added? Explain how this will affect the moles of acid needed to reach the stoichiometric point. 2. During the titration, you added (neutral) DI water from the squirt bottle to wash the acid from the walls of the beaker. Explain whether or not this added water will affect the moles of acid needed to reach the stoichiometric point. How is the calculated Ksp affected? 3. It is not uncommon to add excess acid near the stoichiometric point, so that a larger stoichiometric volume of acid solution is observed in the experiment. Refer to the color pictures of pH indicators at the end of chapter 16 in your textbook. Explain what happens to the color of the solution as the stoichiometric point is passed. Also, explain what will happen to the experimental molar solubility and Ksp if the larger volume is recorded. 4. Suppose the solution in part A was made with tap water instead of DI water. Tap water typically contains what common ion that could affect your results? Explain how the tap water could shift the solubility equilibrium and change your experimental molar solubility. -2+5. Explain how you could use the experimental [OH] in part A to find [Ca]. 2+Also, explain whether [Ca] in part B would be larger or smaller than in part A.
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