1. To prepare a 0.1005 M solution of Ca2+, we need to calculate the mass of CaCl2.2H2O required. The molar mass of CaCl2.2H2O is 147.02 g/mol. The formula shows that there is one Ca2+ ion for every CaCl2 formula unit, so we need to calculate the mass of CaCl2.2H2O required to provide 0.1005 moles of Ca2+.
Mass of CaCl2.2H2O = (0.1005 moles) * (147.02 g/mol) = 14.76851 grams
Therefore, you need to weigh approximately 14.76851 grams of CaCl2.2H2O to prepare 100 mL of a 0.1005 M Ca2+ solution.
2. To prepare a 1 M solution of KCl, we need to calculate the mass of KCl required. The molar mass of KCl is 74.5513 g/mol.
Mass of KCl = (1 mole) * (74.5513 g/mol) = 74.5513 grams
Therefore, you need to weigh approximately 74.5513 grams of KCl to prepare 100 mL of a 1 M KCl solution.
3. The solubility of calcium carbonate in water, recorded in milligrams per liter (mg/L), should be obtained from a reliable source or experiment. Unfortunately, I do not have access to real-time experimental data, so I cannot provide the exact solubility value. Please consult a reputable chemistry reference or conduct an experiment to obtain the solubility value.
4. To calculate the amount of calcium carbonate (CaCO3) not in solution when a tablet with a label claim of 600 mg calcium is dissolved in 250 mL of water, we need to consider the solubility of calcium carbonate.
Assuming the solubility is given in mg/L, we can multiply the solubility by the volume of water to determine the amount of calcium carbonate that can be dissolved:
Amount of calcium carbonate in solution = (Solubility in mg/L) * (Volume of water in L)
Amount of calcium carbonate not in solution = 600 mg - (Amount of calcium carbonate in solution)
5. To increase the solubility of CaCO3 in water, you can add a strong acid, such as hydrochloric acid (HCl). When CaCO3 reacts with an acid, it forms a soluble calcium salt (e.g., calcium chloride) and carbon dioxide gas. The acid provides additional H+ ions, which react with the carbonate ions in CaCO3 to produce CO2 gas, shifting the equilibrium and increasing the solubility of CaCO3.
6. The net ionic equation for the reaction between calcium carbonate and acidified water (in the presence of a strong acid, such as HCl) can be written as follows:
CaCO3(s) + 2H+(aq) → Ca2+(aq) + CO2(g) + H2O(l)
In this equation, CaCO3 reacts with two H+ ions to form soluble Ca2+ ions, carbon dioxide gas, and water.
