Final answer:
The hardness of the water sample with a Ca2+ ion concentration of 1.4×10-3 M is 140.122 ppm of CaCO3.
Step-by-step explanation:
To determine the hardness of a water sample with a given Ca2+ ion concentration, we apply the concept of parts per million (ppm) which is a measure of the concentration of a component by mass relative to the mass of the whole. In this case, we are considering the hardness in terms of CaCO3 equivalent.
Firstly, we convert the molarity of Ca2+ ions (1.4×10-3 M) to its equivalent mass in ppm of CaCO3. Since 1 mole of CaCO3 has a mass of 100.087 g/mol, for every 1 liter (1000 mL) of water having a density of 1.0 g/mL (or 1 kg of water), we calculate the mass of CaCO3 that the given molarity translates to:
According to the formula:
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- Mass of CaCO3 = Molarity of Ca2+ × Molar mass of CaCO3 × Volume of water in liters
We have:
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- Mass of CaCO3 = (1.4×10-3 mol/L) × (100.087 g/mol) × 1 L = 0.140122 g of CaCO3
Now, to convert this mass to ppm, we relate it to mass of water (which is 1000 g since density is 1 g/mL and volume is 1000 mL or 1 L):
Hardness in ppm = (Mass of CaCO3 / Mass of water) × 106 = (0.140122 g / 1000 g) ×106 ppm
Hardness in ppm = 140.122 ppm
Therefore, the hardness of the water sample is 140.122 ppm of CaCO3.