The Ksp of Magnesium Oxalate Abstract The Ksp for the acid catalyzed titration of the saturated oxalate is 1.8 x 10-3. Introduction In this experiment, the solubility equilibrium for the salt magnesium oxalate must be found in order to determine a solubility product constant. Solubility equilibrium is a type of dynamic equilibrium which exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. At the point of equilibrium the solution becomes saturated. The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration.
The specific heat constant for water, 4.184 J/g C, is used for this equation. The specific heat can be found by using The Law of Dulong and Petit: Eq. 3 Cs(aluminum) = slope x 1/atomic weight This equation is used to find specific heat from the graph that will be drawn based on the results of the metal specific heats. II. Materials and Procedure See General Chemistry 101/102 Laboratory Manual (pg.
Results : Reduction Potentials of Several Redox Couples Galvanic Cell Measured Ecell Anode Equation for Anode Reaction Cathode Equation for Cathode Reaction Cu-Zn +1.19 V Zn Zn → Zn2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Mg +3.23 V Mg Mg → Mg2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Fe +0.89 V Fe Fe → Fe2+ + 2e- Cu Cu2+ + 2e- → Cu Zn-Mg +2.00 V Mg Mg → Mg2+ + 2e- Zn Zn2+ + 2e- → Zn Fe-Mg +1.54 V Mg Mg → Mg2+ + 2e Fe Fe2+ + 2e → Fe Zn-Fe +0.29 V Zn Zn → Zn2+ + 2e- Fe Fe2+ + 2e- → Fe Balanced net reaction Cu-Zn Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) Cu-Mg Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) Cu-Fe Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s) Zn-Mg Mg (s) + Zn2+ (aq) → Mg2+ (aq) + Zn (s) Fe-Mg Mg (s) + Fe2+ (aq) → Mg2+ (aq) + Fe (s) Zn-Fe Zn (s) + Fe2+ (aq) → Zn2+ (aq) + Fe (s) Zn-Mg = + 2.00 V, Mg-Cu = + 3.23 V The cell potential of Zn-Cu = 3.23 V – 2.00 V = 1.23 V The measured cell potential of Zn-Cu = +1.19 V The value of the sum of the Zn-Mg and Zn-Cu cell potentials are nearly the same as the Cu-Mg cell potential. Zn-Fe = +0.29 V, Zn-Mg = +2.00 V The cell potential of Fe-Mg = 2.00 V – 0.29 V = 1.71 V The measured cell potential of Fe-Mg = +1.54 V The value of the sum of the Zn-Fe and Zn-Mg cell potentials are nearly the same as the Fe-Mg cell potential. Redox Couple Reduction Potential (measured) Reduction Potential (calculated) % Error Cu2+/ Cu +0.4 V +0.310 V 29% Fe2+/ Fe -0.5 V -0.440 V 14% Zn2+/Zn -0.79 V -0.76 V 4% Mg2+/Mg -2.79 V -2.41 V 16% Calculation for Reduction Potential (V) Ecell = Ecathode - Eanode
Percent H2O in Hydrate is equal 0.34/2.33=14.6% 3. The general formula of barium chloride hydrate is BaClg-nHZO, where n is the number of water molecules. Calculate the theoretical percent water for each value of n—divide the sum of the atomic masses due to the water molecules by the sum of all the atomic masses in the hydrate, and multiply the result by 100. Complete the table. | BaCl2 | BaCl2•H2O | BaCl2•2H2O | BaCl•3H2O | Sum of atomic masses (BaCl2) | 208.23 | 208.23 | 208.23 | 208.23 | Sum of atomic masses (nH2O) | 0 | 18.02 | 36.04 | 54.06 | Sum of atomic masses (hydrate) | 208.23 | 226.25 | 244.27 | 262.29 | Percent water in hydrate (theoretical) | 0% | 7.96% | 14.75% | 20.61% | In this lab we used a Balance, centigram
Conclusion 10 Grams of Potassium chlorate when decomposed produces 3.915576 grams oxygen gas and 6.083363 grams potassium chloride Atomic Weight of Magnesium Introduction In this lab we will determine the atomic weight of magnesium by measuring the amount of hydrogen gas evolved when hydrochloric acid reacts with magnesium. The reaction is as follows: Mg + 2HCl -> H2 + Mg2+ (aq) + 2Cl- (aq) There is a one to one relationship between the number of moles of hydrogen gas evolved and the
There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/
n (3) Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride. co NaOH(s) → Na+(aq) + OH–(aq) ∆H1 = ? Chemistry with Vernier py In this experiment, you will use a Styrofoam-cup calorimeter to measure the heat released by three reactions. One of the reactions is the same as the combination of the other two reactions. Therefore, according to Hess’s law, the heat of reaction of the one reaction should be equal to the sum of the heats of reaction for the other two.
The purpose of the lab was to determine which reactant was the limiting reactant, and to see how much of the other reactant was used. The true molarity of a compound can be defined as the amount of moles per liter of that substance. The equation of this single displacement chemical reaction done during this lab is 2Al(s) + 3CuCl(aq) → 3Cu (s) + 2AlCl2 (aq). In the reaction, the solid Aluminum replaces the Copper in Copper (II) Chloride to produce solid copper, and Aluminum Chloride. In order to find which reactant is the limiting reactant, an equation based on the molarity of the Copper (II) Chloride may be used, or the products of the reaction may be observed.
Part C: Density of an Irregular Shaped Solid 1) Obtain a sample of metal and determine the mass. 2) Fill a 100 mL or 50 mL graduated cylinder with water, recording its volume. 3) Using the water displacement method, determine the volume of the object. 4) Determine the density and percent error. Part D: Density of Methanol 1) Find the mass of an empty 10 mL graduated cylinder, and then fill approximately 9 mL of methanol and record volume.
25.0 cm3 of a 0.10 moldm-3 solution of sodium hydroxide was titrated against a solution of hydrochloric acid of unknown concentration. 27.3 cm3 of the acid was required. What was the concentration of the acid? 3. 25 cm3 of a solution of sodium hydroxide reacts with 15 cm3 of 0.1 mol/dm3 HCl.