A student conducted a back titration to determine the percentage by mass of calcium carbonate in an eggshell. To do this, the student first treated 0.230 g of eggshell with excess acid, using 50.0 cm3 of 0.100 mol dm–3 hydrochloric acid. The calcium carbonate present in the eggshell reacted as follows: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)The excess acid was then titrated using 18.75 cm3 of 0.101 mol dm–3 sodium hydroxide via the following reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Calculate the percentage by mass of calcium carbonate present in the eggshell to 3 significant figures.
Question
A student conducted a back titration to determine the percentage by mass of calcium carbonate in an eggshell. To do this, the student first treated 0.230 g of eggshell with excess acid, using 50.0 cm3 of 0.100 mol dm–3 hydrochloric acid. The calcium carbonate present in the eggshell reacted as follows: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)The excess acid was then titrated using 18.75 cm3 of 0.101 mol dm–3 sodium hydroxide via the following reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Calculate the percentage by mass of calcium carbonate present in the eggshell to 3 significant figures.
Solution
To solve this problem, we need to follow these steps:
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Calculate the amount of HCl that reacted with NaOH. We can use the formula n = c*v, where n is the amount in moles, c is the concentration, and v is the volume. So, n(HCl) = 0.101 mol/dm^3 * 18.75 cm^3 * (1 dm^3/1000 cm^3) = 0.00189 mol.
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This is the amount of excess HCl. The total amount of HCl used was 0.100 mol/dm^3 * 50.0 cm^3 * (1 dm^3/1000 cm^3) = 0.00500 mol. So, the amount of HCl that reacted with CaCO3 is 0.00500 mol - 0.00189 mol = 0.00311 mol.
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From the equation CaCO3 + 2HCl → CaCl2 + CO2 + H2O, we can see that 1 mol of CaCO3 reacts with 2 mol of HCl. So, the amount of CaCO3 that reacted is 0.00311 mol / 2 = 0.00156 mol.
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The molar mass of CaCO3 is 40.08 g/mol (for Ca) + 12.01 g/mol (for C) + 3*16.00 g/mol (for O) = 100.09 g/mol. So, the mass of CaCO3 that reacted is 0.00156 mol * 100.09 g/mol = 0.156 g.
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The percentage by mass of CaCO3 in the eggshell is (0.156 g / 0.230 g) * 100% = 67.8% to three significant figures.
Similar Questions
Plan an experiment to determine the calcium carbonate content in given 2.000 g eggshell sample. You are provided with 2 mol dm-3 Hydrochloric acid, NaOH pellets, distilled water and other lab apparatus. Write a step-by-step process to perform this investigation in the lab starting from NaOH solution preparation to finding the calcium carbonate content in the eggshell.
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