(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a $42.3 \mathrm{~mL}$ aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

First, we need to determine the molarity of the CsOH solution. To do this, we'll follow these steps: 1. Determine the number of moles of CsOH in the solution 2. Calculate the molarity using the volume of the solution Number of moles of CsOH can be calculated by dividing the mass of CsOH by its molar mass: Moles of CsOH = mass / molar mass The mass of CsOH provided is 3.20 g. The molar mass of CsOH (Cs = 132.91 g/mol, O = 16.00 g/mol, H = 1.01 g/mol) is 149.92 g/mol. So the number of moles of CsOH is: moles = (3.20 g) / (149.92 g/mol) = 0.0213 mol Now, use the volume of the solution to calculate the molarity: Molarity = moles / volume The volume of the solution is 25.00 mL, which should be converted to liters: 25.00 mL * (1 L / 1000 mL) = 0.025 L The molarity of the CsOH solution is: Molarity = (0.0213 mol) / (0.025 L) = 0.852 M

Now we need to write the balanced chemical equation for the reaction between the caesium hydroxide (CsOH) and hydroiodic acid (HI) solutions. The reaction is a typical acid-base reaction, where the hydrogen ion (H⁺) from the acid (HI) combines with the hydroxide ion (OH⁻) from the base (CsOH) to form water (H₂O), and the remaining ions form a salt, in this case, caesium iodide (CsI). Here is the balanced chemical equation: CsOH + HI → CsI + H₂O

Given the volume of the CsOH solution required to neutralize an aliquot of the HI solution, we can now calculate the molarity of the HI solution. Since the balanced chemical equation (Step 2) has a 1:1 stoichiometry between CsOH and HI, we can use the volume and molarity of the CsOH solution to determine the moles of HI that reacted and, subsequently, the concentration (molarity) of the HI solution. First, calculate the moles of CsOH in the titration: moles CsOH = molarity * volume We're given that 18.65 mL of the 0.852 M CsOH solution were used, so we convert this volume to L (18.65 mL * (1 L / 1000 mL) = 0.01865 L) and calculate the moles of CsOH: moles CsOH = (0.852 mol/L) * (0.01865 L) = 0.0159 mol Since the stoichiometry between CsOH and HI is 1:1, the moles of HI must be the same as the moles of CsOH: 0.0159 mol Now, use the volume of the HI aliquot to calculate its molarity: molarity HI = moles HI / volume HI The volume of the HI aliquot is given as 42.3 mL, which we convert to L (42.3 mL * (1 L / 1000 mL) = 0.0423 L). Then, calculate the HI molarity: molarity HI = (0.0159 mol) / (0.0423 L) = 0.376 M So the concentration of the hydroiodic acid solution is 0.376 M.