a student determined the Ca^2+ ion content of a 100mL sample of calcium-fortified orange juice. After adding 2.0M HCl solution to the sample, the student filtered the juice to remove the pulp and other particulate matter. Then the tudent added Na2CO3 solution to precipitate CaCO3. The student recovered 0.2818g of CaCO3 from the resulting mixture.
a)why is it important for the student to add HCl before filtering the juice to remove pulp?
i said because if you were to dissolve it with water, the calcium would ppt out as Ca(OH)2 so instead we use HCl so that the Ca forms CaCl2 and stays in solution. is that true?
b)according to the juice label, "each serving contains 20% of the USRDA of calcium," and one serving is defined as 6 oz or 178mL. Does the student's analysis verify this claim? Briefly Explain.
Please be specific and I'd appreciate any help I could get! thanks!
a) Your explanation is partially correct. When calcium ions are present in a solution, they can react with hydroxide ions (OH-) present in water to form calcium hydroxide (Ca(OH)2) which is insoluble and would precipitate out. However, by adding hydrochloric acid (HCl), the hydroxide ions are consumed in a neutralization reaction with the acid, so they are no longer available to react with the calcium ions. Therefore, HCl prevents the formation of calcium hydroxide and keeps the calcium ions in solution.
b) To verify the claim that each serving of the juice contains 20% of the USRDA (U.S. Recommended Daily Allowance) of calcium, we need to compare the amount of calcium recovered in the analysis with the amount of calcium required to meet the 20% claim.
First, let's calculate the amount of CaCO3 recovered in grams per serving:
(0.2818g CaCO3 / 100mL) * 178mL = 0.5008g CaCO3 per serving.
Now, we need to convert CaCO3 to Ca2+ ions since the claim is based on the calcium ion content. To do this, we need to consider the molar mass of CaCO3 (100.09 g/mol) and the molar mass of Ca2+ (40.08 g/mol).
0.5008g CaCO3 * (1 mol CaCO3 / 100.09 g CaCO3) * (1 mol Ca2+ / 1 mol CaCO3) * (40.08 g Ca2+ / 1 mol Ca2+) = 0.2003g Ca2+
Finally, we need to compare the calculated amount of Ca2+ to determine if it meets the 20% claim. Let's calculate:
(0.2003g Ca2+ / 1 serving) * 100% = 20.03% of the USRDA of calcium.
Based on the analysis, the student's result verifies the claim that each serving of the juice contains 20% of the USRDA of calcium. However, it is important to note that any potential sources of error and uncertainties in the analysis might affect the accuracy of this validation.