preview

Na2co3 Lab

Decent Essays

Purpose: Finding the concentration of Ca2+ ions in two samples of hard water. Hypothesis: According to the given molarity of CaCl2 in the table, the concentration of sample one should be 0.400 M, making the hardness of the water 400 mg/L and the concentration of sample four should be 0.100 M, making the hardness of the 40.1 mg/L. Procedures: The first experiment was performed to test the procedure of finding Ca2+ ions in a solution. A known amount of CaCl2 and an excess of Na2CO3 were added to a beaker with deionized water in order to make CaCO3 precipitate. The theoretical yield of CaCO3 was calculated from the mass of the CaCl2. The precipitate was filtered from the solution using the Buchner funnel and aspiration filtration. The precipitate was dried in the oven until the mass stopped decreasing. Then the mass of the funnel with filter was subtracted from the mass funnel, filter and precipitate to find the mass of the dry precipitate. In the tests of samples one and four, the molarity of CaCl2 was approximately known and it was tested using the same process of precipitating CaCO3 by adding excess Na2CO3 and determining experimental molarity of CaCl2 from the mass of that precipitate. After the first experiment to test the …show more content…

mL) x (1 mol Na2CO3 / 1 mol CaCO3) x (1000. mL Na2CO3 / .50 mol Na2CO3) = 4.0 mL, 4.0 mL (1.2) = 4.8 mL (0.200 g x 1 g / 1000. mg) / (.0200 L) x (1/100) = 40.1 mg/L Experimental Water Hardness (with classification) and Percent Yield mg/L = ppm Sample One: (mass of funnel and precipitate 16.37 g)-(mass of funnel 15.48 g)= (mass of CaCO3 precipitate 0.89 g) 0.89 g / .801 g = 110% yield 0.89 g x (1000 mg / (1 g x 20. mL) x (1000mL / 1 L) x (1/100) = 450 mg/L Very hard Sample Two: (mass of funnel and precipitate 15.67 g)-(mass of funnel 15.48 g)= (mass of CaCO3 precipitate 0.19 g) 0.19 g / .200 g = 95% yield 0.19 g x (1000 mg / (1 g x 20. mL) x (1000mL / 1 L) x (1/100) = 95 mg/L Moderately

Get Access