Chemistry Answer Note #24

Question 1 2. Matteus was trying to experimentally determine the solubility of a solute. Could he use each of the following solutions to do this? Respond to each scenario. a. a saturated solution of this same solute b. an unsaturated solution of this same solute c. a supersaturated solution of this same solute d. pure water with no solute added Answer 2. Matteus was trying to experimentally determine the solubility of a solute. Could he use each of the following solutions to do this? Respond to each scenario. a. A saturated solution of this same solute: Yes, Matteus can use a saturated solution of the same solute to experimentally determine its solubility. A saturated solution is one in which the maximum amount of solute has been dissolved in the solvent at a given temperature. By determining the amount of solute that can be dissolved in a given volume of solvent, Matteus can calculate the solubility of the solute. To calculate the solubility of the solute, Matteus can use the formula: Solubility (g/100 mL) = Mass of solute (g)/ Volume of solvent (mL) b. An unsaturated solution of this same solute: Yes, Matteus can also use an unsaturated solution of the same solute to experimentally determine its solubility. An unsaturated solution is one in which the solute has not reached its maximum solubility in the solvent at a given temperature. By adding small amounts of solute and recording the amount that can be dissolved, Matteus can determine the solubility of the solute. c. A supersaturated solution of this same solute: No, Matteus cannot use a supersaturated solution to experimentally determine the solubility of the solute. A supersaturated solution is one in which more solute has been dissolved in the solvent than its maximum solubility at a given temperature. This is achieved by heating the solution to a higher temperature and then cooling it down slowly. Therefore, the solubility of the solute cannot be accurately determined from a supersaturated solution. d. Pure water with no solute added: No, Matteus cannot determine the solubility of the solute from pure water with no solute added. The solubility of a solute is defined as the amount of solute that can be dissolved in a given amount of solvent at a given temperature. In this case, there is no solute present in the pure water, hence there is no solubility to be determined. Question 2 Which of the actions below would increase the solubility of Ag2CrO4? Answer There are several actions that can increase the solubility of Ag2CrO4. These include adding a common ion, increasing the temperature, and increasing the pH. Firstly, adding a common ion, such as chloride (Cl-) or nitrate (NO3-), can decrease the solubility of Ag2CrO4. This is due to the common ion effect, where the addition of a common ion to a solution containing a slightly soluble salt will decrease the solubility of that salt. This is because the common ion will bond with some of the ions of the salt, decreasing the amount of free ions in the solution and reducing the solubility. Secondly, increasing the temperature can also increase the solubility of Ag2CrO4. According to the solubility rules, most solids tend to be more soluble in warmer temperatures. This is due to the increase in kinetic energy and collisions between particles, which can break down the bonds holding the solid together and allow for more ions to be released into the solution. Lastly, increasing the pH can also increase the solubility of Ag2CrO4. As Ag2CrO4 is a slightly basic salt, increasing the pH can convert some of the slightly soluble anions (CrO4 2-) into more soluble hydroxide ions (OH-). This increases the solubility of the salt as more ions are present in the solution. Question 3 determine the mass of 2.5 moles of potassium oxide (K2O) Answer According to the periodic table, the molar mass of potassium is 39.10 g/mol and the molar mass of oxygen is 16.00 g/mol. Since there are two potassium atoms and one oxygen atom in each molecule of potassium oxide (K2O), the molar mass of K2O can be calculated by adding the molar masses of each element together: (2 x 39.10 g/mol) + (1 x 16.00 g/mol) = 94.20 g/mol. To find the mass of 2.5 moles of K2O, we can use the molar mass as a conversion factor. This means multiplying the number of moles by the molar mass. In this case, it would look like: (2.5 moles K2O) x (94.20 g/mol K2O) = 235.5 g K2O. Therefore, the mass of 2.5 moles of potassium oxide is 235.5 grams. Question 4 Which description is not a property of an acid? A) Turns litmus paper red B)Bitter taste C) Corrosive D)Dissolves metals Answer Based on my research, the correct answer is B) Bitter taste. An acid is a chemical compound that donates a proton or accepts an electron pair in reactions. Therefore, it is important to understand the properties of acids in order to properly identify and differentiate them. Let's take a closer look at each of the options provided and determine which one is not a property of an acid. A) Turns litmus paper red This is a property of an acid. Litmus paper is a commonly used indicator to determine if a substance is acidic or basic. An acid will turn litmus paper red, indicating a low pH level. B) Bitter taste This is not a property of an acid. In fact, most acids have a sour taste. It is the base, or alkaline, compounds that have a bitter taste. C) Corrosive This is a property of an acid. Acids have the ability to corrode or eat away at certain materials, causing damage. D) Dissolves metals This is also a property of an acid. When in contact with certain metals, acids can cause a chemical reaction that dissolves the metal. Question 5 What will be the pH of a 0.1 M solution of RbOH Answer According to the Bronsted-Lowry theory, a solution with a pH above 7 is considered to be basic. Therefore, a 0.1 M solution of RbOH, which is a strong base, will have a pH above 7. But to calculate the exact pH of the solution, we need to use the equation pH = log [H+], where [H+] represents the concentration of hydrogen ions in the solution. To calculate the concentration of hydrogen ions, we first need to know the dissociation constant (Kb) for RbOH. Using the formula Kb = [OH-][Rb+]/[RbOH], we can find that the Kb for RbOH is approximately 10^-5. Since RbOH is a strong base, we can assume that it dissociates completely, meaning that the concentration of Rb+ ions will be equal to the concentration of OH- ions. Therefore, in a 0.1 M solution of RbOH, the concentration of OH- ions will also be 0.1 M. Plugging this value into the pH equation, we get pH = -log (0.1) = 1. So, the pH of a 0.1 M solution of RbOH will be 1.