Chemistry Answer Note #25

Question 1 21. Calculate the pH produced when 100.0 mL of solution containing 5.00 g of KOH is mixed with 100.0 mL of a solution containing 6.00 g of HCl? Answer To answer this question, we must first understand what pH is. pH is a measure of how acidic or basic a solution is on a scale of 0 to 14. A pH of 0-6.9 is considered acidic, a pH of 7 is neutral, and a pH of 7.1-14 is basic. In this question, we are mixing two solutions - one containing KOH (a base) and the other containing HCl (an acid). When an acid and a base are mixed together, a neutralization reaction occurs, resulting in the formation of water and a salt. In this case, the salt formed will be KCl. In order to calculate the pH of the resulting solution, we will need to use the equation: pH = -log[H+] First, we need to calculate the number of moles of both KOH and HCl in their respective solutions. This can be done using the formula: moles = mass (in grams) / molar mass. The molar mass of KOH is 56.11 g/mol, and the molar mass of HCl is 36.46 g/mol. For the KOH solution: moles = 5.00 g / 56.11 g/mol = 0.089 moles For the HCl solution: moles = 6.00 g / 36.46 g/mol = 0.164 moles Next, we need to determine which solution will be the limiting reagent, meaning which one will be completely used up during the reaction. To do this, we will use the mole ratio from the balanced chemical equation. The balanced equation for the neutralization reaction between KOH and HCl is: KOH + HCl → KCl + H2O From the equation, we can see that the mole ratio between KOH and HCl is 1:1. This means that for every 1 mole of KOH, we need 1 mole of HCl to react completely. Since we have 0.089 moles of KOH and 0.164 moles of HCl, we can see that HCl is the limiting reagent because it has a higher number of moles. This means that all of the 0.089 moles of KOH will be used up, and there will be some excess HCl left over. To find out how many moles of HCl will be in excess, we will use the following formula: excess moles = initial moles - limiting reagent moles For our equation: excess moles = 0.164 moles - 0.089 moles = 0.075 moles Now that we have determined the excess moles of HCl, we can use this to calculate the concentration of H+ ions in the resulting solution. To do this, we will divide the excess moles by the total volume of the solution (100.0 mL + 100.0 mL), and then by the total number of moles of the resulting solution (0.089 moles of KOH + 0.075 moles of HCl). Concentration of H+ ions = (0.075 moles / 200.0 mL) / (0.089 moles + 0.075 moles) = 0.000375 moles/L Finally, we can plug this concentration into our pH equation to calculate the pH of the resulting solution: pH = -log(0.000375) = 3.45 Therefore, the pH produced when 100.0 mL of solution containing 5.00 g of KOH is mixed with 100.0 mL of a solution containing 6.00 g of HCl is 3.45. The resulting solution will be acidic due to the presence of excess H+ ions from the HCl. Question 2 23. Calculate both [H3O⁺] and [OH¯], to the correct number of significant digits, for each the following solutions. a) pH = -0.55 b) pOH = 9.561 Answer In order to solve this question, we need to first understand the relationship between pH and pOH, and how they relate to the concentrations of hydronium ions ([H3O⁺]) and hydroxide ions ([OH¯]) in a solution. According to the formula for pH, which is defined as the negative logarithm of the hydronium ion concentration, [H3O⁺] = 10^(-pH). Similarly, for pOH, which is defined as the negative logarithm of the hydroxide ion concentration, [OH¯] = 10^(-pOH). a) For the first solution, where pH = -0.55, we can find the [H3O⁺] using the formula mentioned above. [H3O⁺] = 10^(-pH) = 10^0.55 = 0.28 M. The hydronium ion concentration is therefore 0.28 M to the correct number of significant digits. To find the [OH¯] in this solution, we can use the fact that in any solution, the product of [H3O⁺] and [OH¯] is always equal to 1 x 10^(-14) at 25°C. Therefore, [OH¯] = 1 x 10^(14) / [H3O⁺] = 1 x 10^(-14) / 0.28 = 3.6 x 10^(-14) M. The hydroxide ion concentration is therefore 3.6 x 10^(-14) M to the correct number of significant digits. b) For the second solution, where pOH = 9.561, we can use the same formulas as before to find the [H3O⁺] and [OH¯] concentrations. [H3O⁺] = 10^(-pH) = 10^(-9.561) = 2.47 x 10^(-10) M. The hydronium ion concentration is therefore 2.47 x 10^(-10) M to the correct number of significant digits. [OH¯] = 1 x 10^(-14) / [H3O⁺] = 1 x 10^(-14) / 2.47 x 10^(-10) = 4.05 x 10^(-5) M. The hydroxide ion concentration is therefore 4.05 x 10^(-5) M to the correct number of significant digits. Question 3 20. What will be the pH of a 0.1 M solution of RbOH? Answer According to my research, the pH of a 0.1 M solution of RbOH would be approximately 12. The pH scale measures the acidity or basicity of a solution, with 7 being neutral, numbers below 7 indicating acidity, and numbers above 7 indicating basicity. Therefore, a pH of 12 suggests that this solution is highly basic. To understand why this is the case, we must first review the properties of RbOH. RbOH is the chemical formula for rubidium hydroxide, which is a strong base. A strong base completely dissociates in water, meaning all the molecules break apart into their constituent ions. In this case, RbOH will dissociate into Rb+ ions and OH- ions. In a 0.1 M solution of RbOH, there would be 0.1 moles of RbOH per liter of solution. Since it is a strong base, all of these 0.1 moles would dissociate into Rb+ and OH- ions. This means that there would be 0.1 moles of OH- ions in the solution. The pH of a solution can be calculated using the negative logarithm of the concentration of hydrogen ions (H+). In this case, we do not have the concentration of H+ ions, but we do have the concentration of OH- ions. The concentration of H+ ions is equal to the concentration of OH- ions multiplied by the concentration of H+ ions in pure water (1 x 10^-14). Using this information, we can set up the equation: [H+] = (1 x 10^-14)/[OH-]. Plugging in the concentration of OH- ions (0.1), we get [H+] = (1 x 10^-14)/0.1 = 1 x 10^-13. The negative logarithm of 1 x 10^-13 is 13, therefore, the pH of a 0.1 M solution of RbOH is 13. As mentioned earlier, a pH of 13 indicates a highly basic solution. Question 4 A formula for anti-fungal shampoo contains 2.0% W/V ketoconazole. What volume of shampoo, in L, could be made with 68 g of ketoconazole? Answer Ketoconazole is an antifungal medication commonly used in topical treatments for fungal infections such as dandruff and ringworm. In this question, we are given a concentration of 2.0% W/V for ketoconazole in a shampoo formula and asked to calculate the volume of shampoo that can be made with 68 g of ketoconazole. This requires a basic understanding of concentration and conversion between different units of measurement. In order to solve this problem, we can use the following formula: Concentration (C) = Mass of solute (m) / Volume of solution (V) Rearranging the formula, we can solve for the volume of solution (V) using the given concentration and mass of solute: V = m / (C x 100) Substituting the given values, we get: V = 68 g / (2.0 x 100) = 68 / 2 = 34 L Therefore, 34 L of anti-fungal shampoo can be made with 68 g of ketoconazole at a concentration of 2.0% W/V. Question 5 What mass of silver can be plated onto an object in 11.84 minutes at 6.69 A of current? Answer Based on the principles of electroplating and Faraday's laws of electrolysis, the mass of silver that can be plated onto an object in 11.84 minutes at a current of 6.69 A can be calculated using the equation: m = (I*t*M)/(n*F) where: m = mass of silver (in grams) I = current (in amperes) t = time (in seconds) M = molar mass of silver (107.8682 g/mol) n = number of moles of electrons (1 in this case) F = Faraday's constant (96485 C/mol) Using the given values, the calculation would be: m = (6.69 A * 11.84 minutes * 60 seconds/minute * 107.8682 g/mol)/ (1 * 96485 C/mol) = 0.000785 g of silver. Therefore, approximately 0.000785 grams of silver can be plated onto the object in 11.84 minutes at 6.69 A of current.