Question

the calcium in a 300.0-mL sample of natural water was determined by precipitating the cation as CaC204. The precipitate was filtered, washed, and ignited in a crucible with an empty mass of 26.80 g. The mass of the crucible plus CaO (56 g/mol) was 26.90 g. Calculate the concentration of Ca in water in units of gram/100 mL of the water.

Answer 1

Answer on Question #49920 – Chemistry - Other

The calcium in a 300.0-mL sample of natural water was determined by precipitating the cation as $\mathrm{CaC}_2\mathrm{O}_4$. The precipitate was filtered, washed, and ignited in a crucible with an empty mass of 26.80 g. The mass of the crucible plus CaO (56 g/mol) was 26.90 g. Calculate the concentration of Ca in water in units of gram/100 mL of the water.

Solution:

$$\mathrm{Ca}^{2+} + \mathrm{C}_2\mathrm{O}_4^{2-} = \mathrm{CaC}_2\mathrm{O}_4$$ $$\mathrm{CaC}_2\mathrm{O}_4 = \mathrm{CaO} + \mathrm{CO}_2 + \mathrm{CO}$$ $$v(\mathrm{CaC}_2\mathrm{O}_4) = v(\mathrm{CaO}) = v(\mathrm{Ca}^{2+})$$ $$v(\mathrm{CaO}) = \frac{m - m_0}{M_r} = \frac{26.90 - 26.80}{56.08} = 0.0018\ \mathrm{mol}$$ $$m(\mathrm{Ca}^{2+}) = 0.0018\ \mathrm{mol} \times 40.078\ \mathrm{g}/\mathrm{mol} = 0.073\ \mathrm{g}$$ $$c(\mathrm{Ca}^{2+}) = \frac{m}{V(\mathrm{H}_2\mathrm{O})} = \frac{0.073\ \mathrm{g}}{300\ \mathrm{ml}} \times 100 = 0.024\ \mathrm{g}/100\ \mathrm{ml}$$

Answer: 0.024 g /100 mL of the water.

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