Final answer:
To prepare 550 ml of a 0.175 M potassium solution, 6.656 grams of potassium carbonate are needed, taking into account the molar mass of potassium carbonate and the stoichiometry of potassium ions in the compound.
Step-by-step explanation:
To calculate the mass of potassium carbonate (K2CO3) needed to prepare 550 ml of an aqueous solution with a potassium concentration of 0.175 M, we need to consider the formula and molecular weight of potassium carbonate. Potassium carbonate contains 2 moles of potassium ions per mole of compound. Thus, we want to prepare a solution where the total concentration of potassium ions is 0.175 M.
First, we must determine the total moles of potassium ions desired in the solution: Moles of potassium ions = Molarity x Volume (in liters). Moles of potassium = 0.175 mol/L x 0.55 L = 0.09625 moles.
Since there are 2 moles of potassium in each mole of potassium carbonate, we need half the number of moles of potassium carbonate to achieve the desired potassium ion concentration: Moles of K2CO3 = 0.09625 moles / 2 = 0.048125 moles.
We then convert moles of potassium carbonate to grams using its molar mass (138.205 g/mol for K2CO3): Mass of K2CO3 = Moles x Molar mass. Mass of K2CO3 = 0.048125 moles x 138.205 g/mol = 6.656 g. Thus, 6.656 grams of potassium carbonate are required to make 550 ml of a 0.175 M potassium solution.