Find the pOH of a solution containing 5.23 g of HBr in 628 mL of water.
1. Calculate the moles of HBr:
- molar mass of HBr = 1 (H) + 79.9 (Br) = 80.9 g/mol
- moles of HBr = 5.23 g / 80.9 g/mol = 0.065 mol
2. Calculate the molarity of the solution:
- volume of HBr solution = 628 mL = 0.628 L
- molarity = moles / volume = 0.065 mol / 0.628 L = 0.103 M
3. Since HBr is a strong acid, it completely dissociates into H+ and Br- in solution. This means the concentration of H+ in the solution is 0.103 M.
4. Since pOH = -log[OH-] and pKw = 14 at 25°C, we can calculate the pOH as follows:
- pOH = 14 - pH = 14 - (-log[H+])
- pOH = 14 - (-log(0.103)) = 14 - 0.987 = 13.013
Therefore, the pOH of the solution containing 5.23 g of HBr in 628 mL of water is approximately 13.013.