Question

One litre of a sample of hard water contain \(4.44 \mathrm{mg} \mathrm{CaCl}_{2}\) and \(1.9 \mathrm{mg}\) of \(\mathrm{MgCl}_{2}\). What is the total hardness in terms of ppm of \(\mathrm{CaCO}_{3}\) ? (a) \(2 \mathrm{ppm}\) (b) \(3 \mathrm{ppm}\) (c) \(4 \mathrm{ppm}\) (d) \(6 \mathrm{ppm}\)

Short answer

6 ppm

Step-by-step solution

- Calculate Concentration of Calcium

First, convert the concentration of \(\mathrm{CaCl}_2\) from mg to grams (since ppm is mg/L). There is \(4.44\ mg\) of \(\mathrm{CaCl}_2\) in 1 L of water, which equals \(0.00444\ g\) since 1 mg = 0.001 g.

- Calculate Moles of Calcium

Next, calculate the number of moles of calcium. The molar mass of \(\mathrm{CaCl}_2\) is approximately \(40.08\ g/mol \ (\mathrm{Ca}) + 2 \times 35.45 \ g/mol \ (\mathrm{Cl}) = 110.98 \ g/mol\). Therefore, the moles of \(\mathrm{CaCl}_2\) are \(0.00444\ g \div 110.98\ g/mol = 0.000040\ mol\).

- Convert Moles of Calcium to Moles of \(\mathrm{CaCO}_3\)

Assuming a 1:1 molar ratio between \(\mathrm{CaCl}_2\) and \(\mathrm{CaCO}_3\), the moles of \(\mathrm{CaCO}_3\) will also be \(0.000040\ mol\).

- Calculate Mass of \(\mathrm{CaCO}_3\)

Calculate the mass of \(\mathrm{CaCO}_3\) using its molar mass, which is approximately \(40.08\ g/mol (\mathrm{Ca}) + 12.01\ g/mol (\mathrm{C}) + 3 \times 16.00\ g/mol (\mathrm{O}) = 100.09\ g/mol \). Thus, the mass of \(\mathrm{CaCO}_3\) is \(0.000040\ mol \times 100.09\ g/mol = 0.0040036\ g\).

- Convert Mass of \(\mathrm{CaCO}_3\) to ppm

As ppm is equivalent to mg/L, convert the mass of \(\mathrm{CaCO}_3\) from grams to milligrams: \(0.0040036\ g\) is \(4.0036\ mg\). Therefore, the portion due to \(\mathrm{CaCl}_2\) in terms of \(\mathrm{CaCO}_3\) hardness is \(4.0036\ ppm\).

- Repeat Calculation for Magnesium

Perform the same calculation process for \(\mathrm{MgCl}_2\).

Step 6.1 - Calculate Concentration of Magnesium

Convert the concentration of \(\mathrm{MgCl}_2\) from mg to grams: \(1.9\ mg\) is \(0.0019\ g\).

Step 6.2 - Calculate Moles of Magnesium

The molar mass of \(\mathrm{MgCl}_2\) is approximately \(24.31\ g/mol (\mathrm{Mg}) + 2 \times 35.45\ g/mol (\mathrm{Cl}) = 95.21 \ g/mol\). The moles of \(\mathrm{MgCl}_2\) are \(0.0019\ g \div 95.21\ g/mol = 0.00001994\ mol\).

Step 6.3 - Determine Equivalent Moles of \(\mathrm{CaCO}_3\) for Magnesium

Each mole of \(\mathrm{Mg}^{2+}\) ion is equivalent to one mole of \(\mathrm{Ca}^{2+}\) ion due to their equal charge, which means it is also equivalent in hardness to one mole of \(\mathrm{CaCO}_3\). Thus, the moles of \(\mathrm{CaCO}_3\) equivalent is \(0.00001994\ mol\).

Step 6.4 - Calculate Mass of Equivalent \(\mathrm{CaCO}_3\) for Magnesium

Calculate the mass of equivalent \(\mathrm{CaCO}_3\) using its molar mass, \(0.00001994\ mol \times 100.09\ g/mol = 0.0019942\ g\), and convert to milligrams to get \(1.9942\ mg\).

- Calculate Total Hardness

Add the ppm contributions from both calcium and magnesium to find the total hardness in terms of ppm of \(\mathrm{CaCO}_3\): \(4.0036\ mg/L + 1.9942\ mg/L = 5.9978\ mg/L\), which is approximately \(6\ ppm\).

Key concepts

Understanding Parts Per Million (ppm) of CaCO3

When measuring water hardness, scientists and engineers often use the term 'parts per million' (ppm) of calcium carbonate (CaCO3). This unit expresses the concentration of a substance—in this case, CaCO3—within a solution. Imagine you slice up a million pieces of something; if one of those pieces were CaCO3, that would equate to 1 ppm of CaCO3 in the mixture.

To apply this concept to our problem, first, we clarify that 1 ppm is equivalent to 1 milligram of a substance in one liter of water (mg/L). The exercise involves converting the mass of other compounds (CaCl2 and MgCl2) to their equivalent CaCO3 mass, and then expressing this as a concentration. This process is crucial for comparison and treatment purposes in water quality management because it sets a common standard, allowing us to compare hardness levels irrespective of the original compounds present.

Molar Mass Calculation

Molar mass is a fundamental concept in stoichiometry and chemistry at large. It's defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol). To find the molar mass of a compound, such as CaCO3, we sum up the molar masses of its constituent atoms: calcium (Ca), carbon (C), and oxygen (O).

Determining Molar Mass

For our exercise, the molar mass of CaCO3 is calculated by adding the molar masses of one atom of calcium, one of carbon, and three of oxygen. This calculation is crucial for the next step—finding out how many moles of a substance we have based on its mass. With that knowledge, we convert the moles of CaCl2 and MgCl2 to their equivalent in moles of CaCO3, because water hardness is traditionally expressed as if all the hardness-causing ions were in the form of calcium carbonate.

Stoichiometry in Chemistry

Stoichiometry Basics

Stoichiometry is like the recipe book of chemistry; it tells us how much of each reactant is needed to react completely with other reactants and what amount of products we can expect. In the context of our exercise, we use stoichiometry to determine the mole-to-mole conversions between different compounds.

Remember, a mole is a unit of measurement that represents Avogadro's number, 6.022 x 10^23, of anything, just like a dozen represents 12. Stoichiometry allows us to use balanced chemical equations to convert between moles of reactant and product.

Converting Between Compounds

In our solution steps, we assume a 1:1 stoichiometric conversion for the calcium ion in CaCl2 to CaCO3, because both compounds contribute the same calcium ion, which is primarily responsible for the hardness. Similarly, magnesium's contribution to hardness is compared to calcium's by equating the moles of Mg2+ to CaCO3. This is how stoichiometry bridges the gap between different chemical substances and their roles in a chemical process, such as contributing to water hardness.