Question

The concentration of ${\mathrm{Ca}}^{2+}$ in a particular water supply is $5.7\times {10}^{-3}\mathrm{M}$. 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 $20\mathrm{\%}$ of its original level?

Short answer

To reduce the level of ${\mathrm{Ca}}^{2+}$ to $20\mathrm{\%}$ of its original level in $5.0\times {10}^7\mathrm{L}$ of water, the required masses of $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$ are: - Mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ = $315\mathrm{kg}$ - Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ = $1962.3\mathrm{kg}$

Step-by-step solution

Calculate the amount of ${\mathrm{Ca}}^{2+}$ to remove

We begin by finding out how much ${\mathrm{Ca}}^{2+}$ needs to be removed from the water to reduce its concentration to $20\mathrm{\%}$ of its original level. Initially, the concentration of ${\mathrm{Ca}}^{2+}$ is given as $5.7\times {10}^{-3}\mathrm{M}$. To reduce this concentration to $20\mathrm{\%}$ of the original value, we need to find $20\mathrm{\%}\times (5.7\times {10}^{-3}\mathrm{M})$: Final concentration of ${\mathrm{Ca}}^{2+}=0.2\times (5.7\times {10}^{-3}\mathrm{M})=1.14\times {10}^{-3}\mathrm{M}$ Now we calculate the amount of ${\mathrm{Ca}}^{2+}$ to be removed (in moles) from the $5.0\times {10}^7\mathrm{L}$ of water: Amount of ${\mathrm{Ca}}^{2+}$ to remove = (initial concentration - final concentration) $\times$ volume of water $$=(5.7\times {10}^{-3}\mathrm{M}-1.14\times {10}^{-3}\mathrm{M})\times 5.0\times {10}^7\mathrm{L}$$ $$=228\times {10}^4\mathrm{moles}$$

Calculate the moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ required

Now, we will calculate the moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ needed to remove the calculated amount of ${\mathrm{Ca}}^{2+}$. According to the reaction, $\mathrm{Ca}(\mathrm{OH}{)}_2$ reacts with ${\mathrm{HCO}}_3^{-}$ in a $1:2$ ratio: $$\mathrm{Ca}(\mathrm{OH}{)}_2+2{\mathrm{HCO}}_3^{-}\to {\mathrm{CaCO}}_3\downarrow +2{\mathrm{H}}_2\mathrm{O}$$ Since there are $1.7\times {10}^{-3}\mathrm{M}$ of ${\mathrm{HCO}}_3^{-}$ in the water, we have: Moles of ${\mathrm{HCO}}_3^{-}$ = $1.7\times {10}^{-3}\mathrm{M}\times 5.0\times {10}^7\mathrm{L}=85\times {10}^4\mathrm{moles}$ Thus, for every mole of $\mathrm{Ca}(\mathrm{OH}{)}_2$ we need two moles of ${\mathrm{HCO}}_3^{-}$. Therefore, we can calculate the moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ required: Moles of $\mathrm{Ca}(\mathrm{OH}{)}_2$ = $\frac{1}{2}\times 85\times {10}^4\mathrm{moles}=42.5\times {10}^4\mathrm{moles}$

Calculate the masses of $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$

First, we will find the mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ required: Mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ = moles $\times$ molar mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ Using the molar mass of \(\mathrm{Ca(OH)_2 = 74.1\,\mathrm{g/mol}\), we get: Mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ = $42.5\times {10}^4\mathrm{moles}\times 74.1\mathrm{g}/\mathrm{mol}=315\mathrm{kg}$ Next, we find the moles of remaining ${\mathrm{Ca}}^{2+}$, which will react with ${\mathrm{Na}}_2{\mathrm{CO}}_3$: Moles of remaining ${\mathrm{Ca}}^{2+}$ = $228\times {10}^4-42.5\times {10}^4=185.5\times {10}^4\mathrm{moles}$ According to the reaction, ${\mathrm{Ca}}^{2+}$ reacts with ${\mathrm{Na}}_2{\mathrm{CO}}_3$ in a $1:1$ ratio. Thus, the moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ required is the same as the moles of remaining ${\mathrm{Ca}}^{2+}$: Moles of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ = $185.5\times {10}^4\mathrm{moles}$ Now, we calculate the mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$: Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ = moles $\times$ molar mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ Using the molar mass of \(\mathrm{Na_2CO_3 = 106\,\mathrm{g/mol}\), we get: Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ = $185.5\times {10}^4\mathrm{moles}\times 106\mathrm{g}/\mathrm{mol}=1962.3\mathrm{kg}$

Present the final results

To reduce the level of ${\mathrm{Ca}}^{2+}$ to $20\mathrm{\%}$ of its original level in $5.0\times {10}^7\mathrm{L}$ of water, the required masses of $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$ are: - Mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ = $315\mathrm{kg}$ - Mass of ${\mathrm{Na}}_2{\mathrm{CO}}_3$ = $1962.3\mathrm{kg}$

Key concepts

Understanding Stoichiometry in Chemical Reactions

Stoichiometry is a fundamental concept in chemistry that relates to the quantitative relationship between reactants and products in a chemical reaction. It is the calculation of the quantities of chemical elements or compounds involved in chemical reactions. To grasp why stoichiometry is crucial in the exercise above, imagine you are cooking and need to adjust a recipe for a larger number of guests. You'll need to proportionally increase the amount of each ingredient to maintain the taste; likewise, in chemistry, stoichiometry helps us adjust the reactants to get the desired amount of product.

To solve the problem in the exercise, you need to understand the stoichiometry of the reaction between calcium hydroxide $\mathrm{Ca}(\mathrm{OH}{)}_2$ and bicarbonate ion ${\mathrm{HCO}}_3^{-}$, which follows a 1:2 molar ratio. This ratio dictates the proportions in which reactants combine to form products, ensuring that the chemical reaction is balanced and enabling us to calculate the exact amount of $\mathrm{Ca}(\mathrm{OH}{)}_2$ needed to treat the water hardness.

Determining Molar Concentration in Solutions

Molar concentration, often referred to as molarity, is another core concept highlighted in the exercise. It indicates the number of moles of a solute per liter of solution, expressed in moles per liter (M). For example, if we know the molar concentration of ${\mathrm{Ca}}^{2+}$ ions in water, we can calculate the moles of ${\mathrm{Ca}}^{2+}$ in any volume of water, which is essential for water hardness treatment.

Understanding molar concentration is fundamental to solving the problem because you need to know how much ${\mathrm{Ca}}^{2+}$ is present initially, how much you want to remove, and the necessary quantities of $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$ to reach the desired concentration. The concept of molarity is applied throughout the step-by-step solution to figure out the moles of substances, and naturally extends to calculating masses when multiplied by the respective molar masses.

Balancing Chemical Equations for Substance Calculation

The balancing of chemical reactions is a critical component of stoichiometry and is necessary to ensure that the law of conservation of mass is respected. This means the number of atoms for each element must be the same on both sides of the reaction equation. For instance, in the exercise, the reaction between $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{HCO}}_3^{-}$ is balanced with a 1:2 ratio. Without a balanced equation, stoichiometric calculations would be meaningless because they rely on a correct understanding of how molecules interact.

By ensuring that the equation is balanced, as we see in the given solution, you can then use the established ratios of reactants to products to calculate the correct amounts of reagents needed to treat the water. The balanced equation not only illustrates the law of conservation of mass but also enables us to apply the stoichiometric coefficients to determine the mass of $\mathrm{Ca}(\mathrm{OH}{)}_2$ and ${\mathrm{Na}}_2{\mathrm{CO}}_3$ required in the treatment process.