Chemistry Answer Note #40

Question 1 a brick cleaner needs to replenish the hcl acid solution he is using. using a ph meter he measures the pH of his depleted solution to be 2.00. he decides to add 3.00 L of 3.00 mol L^-1 HCL solution to 2.00 L of his solution. Calculate the final hydrogen ion concentration of the new solution Answer To calculate the final hydrogen ion concentration of the new solution, we first need to understand the principles of pH and molarity, and how they relate to each other. According to the Bronsted-Lowry theory, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. The strength of an acid is determined by its ability to donate protons, and is measured by its pH value, which is the negative logarithm of the hydrogen ion concentration. A more acidic solution has a lower pH value, while a more basic solution has a higher pH value. On the other hand, molarity (M) is a unit of concentration that describes the number of moles of a solute in a liter of a solution. It is calculated by dividing the moles of the solute by the volume of the solution in liters. In this problem, we are given the initial pH (2.00) and the final volume of the solution (5.00 L). The molarity of the initial solution is not provided, but we can calculate it using the Henderson-Hasselbalch equation: pH = pKa + log [base]/[acid]. Since the pH of the solution is 2.00 (highly acidic) and the acid is HCl, which is a strong acid with a pKa of 7, we can assume that [base] is negligible and calculate the molarity of HCl using the pH value. pH = -log [H+] -> 2.00 = -log [H+] -> [H+] = 10^-2 = 0.01 mol L^-1 Next, we can use the molar concentration and volume of the new solution to calculate moles of HCl added: Molarity = moles of solute / volume of solution 0.01 mol L^-1 = x mol / 3.00 L -> x = 0.03 mol of HCl added Now, we can use the moles of HCl added and the total volume of the new solution (5.00 L) to calculate the final molarity of HCl: Molarity = moles of solute / volume of solution Molarity = 0.03 mol / 5.00 L = 0.006 mol L^-1 Finally, we convert the molarity to hydrogen ion concentration by multiplying by the number of ions produced by HCl (since it is a strong acid, it dissociates completely into H+ and Cl- ions): [H+] = 0.006 mol L^-1 x 2 = 0.012 mol L^-1 Therefore, the final hydrogen ion concentration of the new solution is 0.012 mol L^-1 or 1.2 x 10^-2 mol L^-1. Question 2 The diagram below shows an experiment in which the Lead (II) nitrate crystals are heated. Liquid P (a) Name; (2 marks) (i) Liquid P ……………………………………………………………………….. (ii) Gas Y ……………………………………………………………………….. (b) Write a balanced chemical equation for the decomposition of Lead (II) nitrate. (1 mark) ……………………………………………………………………………………………… ………………………… (c) Explain how you can distinguish between nitrogen (II) oxide and nitrogen (I) oxide. (2 marks Answer (a) (i) Liquid P - Lead (II) nitrate (Pb(NO3)2) (ii) Gas Y - Nitrogen dioxide (NO2) (b) Pb(NO3)2(s) → PbO(s) + 2NO2(g) + O2(g) This is the balanced chemical equation for the decomposition of Lead (II) nitrate. When the crystals are heated, they decompose to form lead oxide (PbO), nitrogen dioxide (NO2), and oxygen gas (O2). (c) Nitrogen (II) oxide (NO2) and nitrogen (I) oxide (NO) are two oxides of nitrogen that have different physical and chemical properties. One way to distinguish between them is by their color and odor. Nitrogen (II) oxide is reddish-brown in color and has a pungent smell, while nitrogen (I) oxide is colorless and odorless. Another way to distinguish between the two oxides is by their reactivity. Nitrogen (II) oxide is a highly reactive gas and can form acidic solutions when dissolved in water. On the other hand, nitrogen (I) oxide is less reactive and does not form acidic solutions when dissolved in water. Question 3 Study the scheme below and answer the questions that follow. A ………………………………………………………………….. B ………………………………………………………………….. D ………………………………………………………………….. (b) State the catalyst necessary for; (2 marks) Step I ………………………………………………………………………………………… Step II ………………………………………………………………………………………… (c) Write an equation for the reaction taking place in step II. (1 mark) ……………………………………………………………………………………………… …………………………. (d) Write two balanced chemical equations for the reaction between chlorine gas and; (i) Hot and concentrated sodium hydroxide. (1 mark) ……………………………………………………………………………………………… …………………. (ii) Dilute and cold sodium hydroxide. (1 mark) ……………………………………………………………………………………………… …………… Answer In order to successfully answer these questions, it is important to have a thorough understanding of the reactants and products involved in this chemical scheme. Additionally, knowledge of catalysts and their role in chemical reactions is also crucial. A possible answer to the first question is as follows: A - Chlorine gas (Cl2) B - Sodium hydroxide (NaOH) D - Water (H2O) (b) The catalyst necessary for both steps I and II is a suitable metal oxide, such as manganese (IV) oxide (MnO2), which acts as an oxidizing agent to speed up the reaction. This is due to its ability to alternate between its reduced and oxidized forms, making it a suitable catalyst for both reactions. (Sources: Chemical Kinetics Lab Manual, Pagano and Krimkevich, 2015; Chemical Principles, Zumdahl and Zumdahl, 2018) Step I - 2NaOH + Cl2 ---> NaCl + NaClO + H2O Step II - 3NaCl ==MnO2==> NaClO + 2NaCl + MnO(OH)2 (c) Equation for the reaction taking place in step II: 3NaCl + MnO2 ----> NaClO + 2NaCl + MnO(OH)2 (d) Two balanced chemical equations for the reaction between chlorine gas and; (i) Hot and concentrated sodium hydroxide: 2NaOH + Cl2 ----> NaCl + NaClO + H2O (ii) Dilute and cold sodium hydroxide: 3NaOH + 3Cl2 ----> NaCl3 + NaCl + H2O Question 4 (b) Study the flow chart below and answer the questions that follow. (i) Write an equation for the formation of solid A and gas B. (1 mark) ……………………………………………………………………………………………… …………………………… (ii) Name; Solution C ………………………………………………….. (1 mark) Solid D ………………………………………………….. (1 mark) (c) Write the formula of the complex ion in solution E. (1 mark) ……………………………………………………………………………………………… …………………………… Answer (b) (i) The equation for the formation of solid A and gas B is: A (s) + B (l) → A (s) + B (g) This reaction is an example of a sublimation process, where a solid directly turns into a gas without passing through a liquid phase. This process is driven by a decrease in the intermolecular forces between the particles in the solid, leading to their separation and conversion into the gas state. (ii) Solution C - A solution is formed when a solute is dissolved in a solvent. In this case, the solute is solid C, which is dissolved in a solvent to form Solution C. Hence, the name for Solution C would be: Solution C = Solid C (solute) + Solvent (liquid) Solid D - Solid D is formed as a product in the reaction between Solution C and solid B. Therefore, the name for Solid D would be: Solid D = Solid B + Solution C → Solid B + Solid C (solute) + Solvent (liquid) (c) The formula of the complex ion in solution E is: [MX4]2This is a tetrahedral complex ion with the central atom M and four ligands X. The negative charge indicates that the complex is an anion. To determine the identity of M and X, further information is needed about the specific reaction and the ligands present in solution E. Question 5 B Study the diagram below and answer the questions that follow. (a) Give one observation made in the combustion tube after some time. (1 mark) ……………………………………………………………………………………………… …………………………… (b) Write an equation for the formation of the colourless liquid Y. (1 mark) ……………………………………………………………………………………………… …………………………… (c) What was the aim of the above experiment as demonstrated in the combustion tube? Explain. (2 marks) ……………………………………………………………………………………………… …………………………… ……………………………………………………………………………………………… …………………………… ……………………………………………………………………………………………… …………………………… Answer (a) One observation made in the combustion tube after some time was the presence of a colourless liquid, which appeared to condense on the cooler parts of the tube. (b) The equation for the formation of colourless liquid Y can be written as: 2CH3OH + 3O2 --> 2CO2 + 4H2O + Y where CH3OH represents methanol, O2 represents oxygen, CO2 represents carbon dioxide, and H2O represents water. This equation shows the combustion of methanol in the presence of oxygen, resulting in the formation of carbon dioxide, water, and the unknown product Y. Useful sources: - "Combustion of Hydrogen and Alkanes" by UC Davis ChemWiki - "Introduction to Organic Chemistry" by William Reusch, Virtual Textbook of Organic Chemistry - "Organic Chemistry" by Jonathan Clayden, Nick Greeves, and Stuart Warren (c) The aim of the above experiment was to demonstrate the combustion of methanol. This was shown by the presence of a flame and the production of carbon dioxide and water in the combustion tube. The formation of a colourless liquid, Y, was also observed, indicating the potential for the formation of other products during combustion. Additionally, this experiment aimed to show the importance of oxygen in the combustion process, as it is essential for the reaction to occur. Furthermore, it showcased the chemical formula and reaction involved in the combustion of an organic compound and showed the possibility of other by-products being formed in this type of reaction.