Chemical Equilibrium: Le Chatelier Principle By Sarah Ramos and Kristina Todorovic Chemistry 203 DEN Dr. Mohamed El-Maazawi Part A. Acid-Base Indicators Purpose In this part of the experiment‚ we will find a reagent that will shift the acid-base equilibrium reaction described by Equation (2) in one direction and then a second reagent that will cause the equilibrium position to shift back in the opposite direction. Introduction An acid–base indicator
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titration experiments had to be done. In the first series‚ a titration is completed of sodium hydroxide (NaOH) with KHP. Three reps were completed and the resulting data led to the finding of the molarity of the sodium hydroxide. The second part of the titration series involved titrating the same prepared sodium hydroxide solution with Hydrochloric Acid (HCL). This is how the molarity of the HCL was determined. The last part to the titration series was necessary to determine the equivalent weight
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Investigating Chemical Equilibrium Date: 30th April 2013 Due Date: 15th May 2013 Prepared For: M. Seraji Prepared by: Andrea Odunze Abstract Many reactions proceed to a state of equilibrium. A chemical reaction at equilibrium‚ where the rates of the forward reaction and reverse reaction are equal‚ looks like this: A + B AB There are three factors‚ according to Le Chatelier’s principle‚ that affect the equilibrium position and equilibrium constant. These are the concentrations of products
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Buffer Post Lab Calculation 1. We added too much NaOH to our buffer 1 solution. The original pH was 5.32‚ with 10mL of buffer and 10mL of DI water. We added 5mL of NaOH and the pH changed to 12.12. If 5 mL changed the pH by about 7‚ then we predict that about 2 mL of NaOH would change the pH by about 2. This prediction is based on the fact that 2.5mL is half of 5mL‚ and 3.5 is half of 7‚ so adding a little less than 2.5mL would give us a difference of about 2. For the buffer 2 solution‚ the original
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reaction below. 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) + HCl(aq) (in excess) (limiting reagent) (1) (unreacted) This reaction cannot be used directly to titrate the CaCO3 because it is very slow when the reaction is close to the endpoint. Instead the determination is achieved by adding an excess of hydrochloric acid to react with all of the CaCO3 and then titrating the remaining unreacted HCl with NaOH solution to determine the amount of acid which did not react with the
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Strong or Weak Acid/Base pH 0.1 M HCl (aq) 0.1 M HC2H3O2 (aq) 0.1 M NaOH (aq) 0.1 M NH3 (aq) Question 1. Both acids above have the same concentration‚ 0.1 M‚ but they exhibit different pH value. Explain. Question 2. Both bases above have the same concentration‚ 0.1 M‚ but pH of NH3(aq) is lower than that of NaOH. Explain. 7 III. a) Observation of HCl/NaOH product. b) Identity of the product: c) Write the reaction occurred between HCl and NaOH. __________________________________________________________________________
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and its (conjugate acid) Carbonic acid was used. The purpose of the experiment was to test the capacity of an undiluted and diluted buffer solution as well as examining the buffering capacity of distilled water using measured concentrations of NaOH and HCL. These replacements of strong acids and bases for weaker ones give buffers their ability to moderate pH. (Stoker 2013). Part 2 of the experiment‚ the buffering capacity of lake water was tested. The ions naturally present in rivers are buffering
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Trials of HCl and 1 Trial of Acetic Acid (Vinegar) Trial # | Va | Vb | 1 | 10.00 mL | Start:0.000 mLFinish:3.700 mLVb:3.700 mL | 2 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | 3 | 10.00 mL | Start:0.000 mLFinish:3.750 mLVb:3.750 mL | Vinegar | 5.00 mL | Start:0.000 mLFinish:12.100 mLVb:12.100 mL | Figure 1 Calculations Involved With Experiment Mb Value using 0.1 M HCL and measured Va and VbTrial 1:(0.1)(10.00)=(3.700)(x) x=0.270 | Mb Value using 0.1 M HCL and measured
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3.13 Test for Saponin glycosides 2g of dried plant powder add 5ml of distill water shake vigorously. Persistent of foam indicate the presence of saponin glycosides. 3.14 Test for Steroids (Salkowski’s test) 5ml of test solution add with 2ml of Chloroform and then Conc. H2SO4 sides of the test tubes. The appearance of color at the interface indicates the presence of steroids. 3.15 Experimental Procedure for preliminary Phytochemical studies (Brindha et al.‚ 1981) S. No Name of the Test
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of H2PO4 buffer solution with 0.1 M HCl and 0.1 M NaOH. The volume of 0.1 M HCl is about double of the amount 0.1 M NaOH used to lower/raise the pH of a blood buffer. In this experiment‚ HCl (a strong acid) and NaOH (a strong base) are used as examples of strong acids/bases‚ and the titration with H2PO4 shows the effect on a buffer solution. The assumption was the addition of large amounts of HCl will lower the pH‚ while the addition of large amounts of NaOH will increase the pH‚ while small amounts
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