Question

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} M .\) The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

Short answer

To reduce the level of Ca²⁺ to 20% of its original level in the given water supply, \(1.69 \times 10^{7} \mathrm{g}\) of Ca(OH)₂ and \(4.5 \times 10^{6} \mathrm{g}\) of Na₂CO₃ must be added.

Step-by-step solution

Analyze the target concentration for Ca²⁺

We are told to reduce the Ca²⁺ concentration to 20% of its original level. The original concentration is \(5.7 \times 10^{-3} M\). So, the target concentration of Ca²⁺ will be: \[C_{Ca^{2+}} =0.20 \times 5.7 \times 10^{-3} \mathrm{M} = 1.14 \times 10^{-3} \mathrm{M}\]

Find the amount of Ca²⁺ that needs to be removed

To find how much Ca²⁺ concentration needs to be removed, subtract the target concentration from the initial concentration: \[\Delta C_{Ca^{2+}} = 5.7 \times 10^{-3} \mathrm{M} - 1.14 \times 10^{-3} \mathrm{M} = 4.56 \times 10^{-3} \mathrm{M}\]

Analyze the amounts of each substance in moles

Calculate the moles of calcium ions (Ca²⁺), bicarbonate ions (HCO₃⁻), calcium hydroxide (Ca(OH)₂), and sodium carbonate (Na₂CO₃) needed for this change: \[\text{moles of } Ca^{2+} = \Delta C_{Ca^{2+}} \times V = 4.56 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 2.28 \times 10^{5} \mathrm{mol}\] \[\text{moles of } HCO_3^{-} = C_{HCO_3^-} \times V = 1.7 \times 10^{-3} \mathrm{M} \times 5.0 \times 10^{7} \mathrm{L} = 8.5 \times 10^{4} \mathrm{mol}\] Since, one mole of Ca(OH)₂ reacts with one mole of Ca²⁺, we have: \[\text{moles of } Ca(OH)_2 = \text{moles of } Ca^{2+} = 2.28 \times 10^{5} \mathrm{mol}\] As we know, one mole of Na₂CO₃ reacts with two moles of HCO₃⁻, so we get: \[\text{moles of } Na_2CO_3 = \frac{1}{2} \times \text{moles of } HCO_3^{-} = \frac{1}{2} \times 8.5 \times 10^{4} \mathrm{mol} = 4.25 \times 10^{4} \mathrm{mol}\]

Calculate the masses of Ca(OH)₂ and Na₂CO₃

With the moles of each substance determined, calculate the masses of Ca(OH)₂ and Na₂CO₃ needed: \[\text{mass of } Ca(OH)_2 = \text{moles of } Ca(OH)_2 \times M_{Ca(OH)_2} = 2.28 \times 10^{5} \mathrm{mol} \times 74.09 \frac{\mathrm{g}}{\mathrm{mol}} = 1.69 \times 10^{7} \mathrm{g}\] \[\text{mass of } Na_2CO_3 = \text{moles of } Na_2CO_3 \times M_{Na_2CO_3} = 4.25 \times 10^{4} \mathrm{mol} \times 105.99 \frac{\mathrm{g}}{\mathrm{mol}} = 4.5 \times 10^{6} \mathrm{g}\] Thus, we need to add \(1.69 \times 10^{7} \mathrm{g}\) of Ca(OH)₂ and \(4.5 \times 10^{6} \mathrm{g}\) of Na₂CO₃ to the water to achieve the desired reduction in Ca²⁺ concentration.

Key concepts

Calculating Molar Concentration

Understanding molar concentration is fundamental in chemistry and pivotal for tasks such as reducing water hardness. Molar concentration, often represented by the symbol 'M', indicates the moles of a substance per liter of solution. It's a way of expressing how much of a solute is present in a given volume of solvent.

For example, when given a concentration of \(5.7 \times 10^{-3} M\) for calcium ions (\(\mathrm{Ca}^{2+}\)), we can interpret this to mean we have 0.0057 moles of \(\mathrm{Ca}^{2+}\) for every liter of water. When reducing water hardness, understanding this concept helps us calculate how much of a reacting agent we need to introduce to achieve the desired concentration level of ions in the water.

To calculate the target concentration, we simply multiply the initial molar concentration by the percentage of the reduction desired—in this case, 20%, leading to the target concentration formula: \[ C_{Ca^{2+}} = 0.20 \times \text{Initial Concentration} \]. This step is crucial for determining the amount of reactive agent needed for the treatment process.

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of the mole. It allows us to predict how much reactants are required to produce a certain amount of product, or conversely, how much product can be formed from a certain amount of reactants.

In water hardness treatment, for instance, stoichiometry helps us understand that a precise ratio of calcium hydroxide (\(\mathrm{Ca(OH)_2}\)) is needed to react with the present calcium ions to reduce hardness. Similarly, sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3}\)) is used to react with bicarbonate ions. Since one mole of \(\text{Ca(OH)_2}\) reacts with one mole of \(\text{Ca}^{2+}\), and one mole of \(\text{Na}_2\mathrm{CO}_3\) reacts with two moles of \(\text{HCO}_3^-\), stoichiometry allows us to calculate the exact amounts needed for these reactions to happen efficiently without excess reactant.

Water Treatment Chemistry

Water treatment chemistry involves various processes and reactions to remove impurities and make water suitable for a specific purpose, like consumption, industrial use, or releasing back into the environment. Hard water, which contains high levels of minerals such as calcium and magnesium, can cause problems like scaling in pipes and reduced effectiveness of soap.

To soften hard water, chemical processes such as the addition of calcium hydroxide or sodium carbonate are used, which react with the calcium ions and bicarbonate ions to form precipitates that can be removed from the water. The exercise described utilizes these principles to calculate the precise amounts of chemicals needed to achieve a desired level of water hardness reduction.

Water treatment professionals rely on these calculations and the stoichiometry of the reactions involved to ensure they add the correct amounts of chemicals. This maintains the efficiency and safety of the water treatment process, ensuring that the treated water meets regulatory standards and is fit for its intended use.