Question

A particular water sample contains 56.9 ppm \(\mathrm{SO}_{4}^{2-}\) and \(176 \mathrm{ppm} \mathrm{HCO}_{3}^{-},\) with \(\mathrm{Ca}^{2+}\) as the only cation. (a) How many parts per million of \(\mathrm{Ca}^{2+}\) does the water contain? (b) How many grams of \(\mathrm{CaO}\) are consumed in removing \(\mathrm{HCO}_{3}^{-}\), from \(602 \mathrm{kg}\) of the water? (c) Show that the \(\mathrm{Ca}^{2+}\) remaining in the water after the treatment described in part (b) can be removed by adding \(\mathrm{Na}_{2} \mathrm{CO}_{3}.\) (d) How many grams of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) are required for the precipitation referred to in part (c)?

Short answer

The water contains 23.7 ppm of Ca2+, 660 g of CaO are consumed in removing HCO3- from 602 kg of the water, the remaining Ca2+ in the water can be removed by adding Na2CO3, and 1418 g of Na2CO3 are required for the precipitation.

Step-by-step solution

Calculate Ca2+ in ppm

The molar masses of SO4^2- and Ca2+ are 96 g/mol and 40 g/mol respectively. The ppm (parts per million) is a unit of concentration. Assuming that SO4^2- and Ca2+ are in a 1:1 ratio (because CaSO4 is formed), the concentration of Ca2+ would be \(Ca^{2+} = \frac{56.9 \times 40}{96} = 23.7 ppm\).

Calculate amount of CaO for removing HCO3-

The formation of CaCO3 from CaO and HCO3- is in a 1:1 ratio. Therefore, for each mole of HCO3-, one mole of CaO is needed. Given that the molar mass of CaO is 56 g/mol and the ppm of HCO3- is 176 (which means 176 mg/kg), we have: \[CaO = \frac{176 \times 602 \times 56}{96} = 660 g\] of CaO consumed in removing HCO3- from 602 kg of water.

Prove that remaining Ca2+ can be removed by Na2CO3

The reaction between Ca2+ and Na2CO3 leads to formation of CaCO3 and 2Na+. This reaction proves that the remaining Ca2+ in the water can be precipitated out by adding Na2CO3.

Calculate required amount of Na2CO3

Again using stoichiometry, for each mole of Ca2+ one mole of Na2CO3 is needed. Given that the molar mass of Na2CO3 is 106 g/mol and the ppm of Ca2+ left in the water after removal of HCO3- is 23.7 (which means 23.7 mg/kg), we have: \[Na2CO3 = \frac{23.7 \times 602 \times 106}{40} = 1418 g\] of Na2CO3 required to precipitate the remaining Ca2+.

Key concepts

Calcium Ion Removal

Calcium ions are prevalent in water sources and their removal is essential for preventing scale formation and ensuring water safety. The removal of calcium ions from water involves chemical reactions that precipitate these ions out of the solution.
In the provided exercise, calcium ions (\(\mathrm{Ca}^{2+}\)) are balanced with sulfate ions (\(\mathrm{SO}_{4}^{2-}\)) in a 1:1 molar ratio. This results in the formation of calcium sulfate, which allows us to determine the concentration of calcium ions using stoichiometry based on the sulfate concentration.
Calculating calcium ion concentration in parts per million (ppm) involves understanding this balanced relationship. Here, with 56.9 ppm of sulfate ions, we apply a stoichiometric calculation using the respective molar masses of the ions:

• Molar mass of \(\mathrm{SO}_{4}^{2-}\): 96 g/mol
• Molar mass of \(\mathrm{Ca}^{2+}\): 40 g/mol

Thus, the concentration of calcium ions is 23.7 ppm, calculated by \(\text{Ca}^{2+} = \frac{56.9 \times 40}{96}\), demonstrating the simplicity of stoichiometric calculations in determining ions in water.

Sulfate Concentration

Sulfate concentration in water is an important parameter as it often influences the overall chemistry and potential reactions within the water. In the given problem, the sulfate concentration is the starting point for understanding the composition of the solution and its reaction potentials.
We know that sulfate ions contribute to the total ionic balance of the water, typically balanced by a corresponding cation. In this problem, the lone cation balancing sulfate is the calcium ion. This information helps establish a foundation for calculating concentrations of other ions like calcium using stoichiometry.
Given that the sulfate concentration here is 56.9 ppm, it lends itself to being directly related to how much calcium sulfate could potentially precipitate out of the water, as seen in the conversion of sulfate ppm to calcium ppm. Understanding this allows us to predict water chemistry reactions and treatment needs appropriately, ensuring precise control over water quality and treatment processes.

Bicarbonate Interaction

Bicarbonate ions play a critical role in water chemistry, often acting as indicators of the water's potential alkalinity and support chemical reactions. In the given exercise, the focus is on removing bicarbonate ions \(\mathrm{HCO}_{3}^{-}\) using calcium oxide (CaO).
When \(\mathrm{CaO}\) reacts with bicarbonate ions, it ultimately facilitates the removal of bicarbonates by precipitating them as calcium carbonate (\(\mathrm{CaCO}_{3}\)). This reaction is pivotal as it helps manage the bicarbonate level thereby lowering the water's alkalinity.
The process involves a 1:1 molar reaction ratio, indicating that each mole of bicarbonate requires one mole of CaO for complete removal. In the context of the provided scenario:

• Bicarbonate concentration is 176 ppm (mg/kg of water).
• The required amount of \(\mathrm{CaO}\) for treat 602 kg of water is determined by using its molar mass, engaging a simple stoichiometric equation:
• \( \text{CaO} = \frac{176 \times 602 \times 56}{96} = 660 \, \text{g} \)

These calculations give precise measurements necessary for effectively managing bicarbonate interactions in water treatment.

Stoichiometry Calculations

Stoichiometry is the cornerstone of determining the amounts of reactants and products in chemical reactions, especially in water chemistry. It allows for precise calculations of substances required or produced.
In the problem, several stoichiometric calculations are used to orchestrate the removal of calcium ions and bicarbonates. By understanding the relationships between the reactants, we can solve real-life chemistry problems.

• Calcium Ion Removal: To find the \(\text{ppm}\) of \(\text{Ca}^{2+}\) we used the ratio dictated by its reaction with sulfates.
• Sulfate Concentration: Sulfate ions dictated how much calcium ion is in the solution as they form a 1:1 ratio when bound in \(\text{CaSO}_4\).
• Bicarbonate Interaction: The 1:1 molar reaction ratio of \(\text{CaO}\) to bicarbonate ions facilitated the removal of bicarbonates.
• Na2CO3 Precipitation: The remaining \(\text{Ca}^{2+}\) ions were calculated for removal by \(\text{Na}_2 ext{CO}_3\), again using a 1:1 molar ratio to ensure full precipitation.

Essentially, stoichiometry ensures chemical reactions are balanced and expected results are achieved, allowing us to perform accurate water treatment, achieving desired levels of purity.