Problem 46 - Chemistry The Science in Context

Question

In an alkalinity titration of a \(100.0 \mathrm{mL}\) sample of water from a hot spring, \(2.56 \mathrm{mL}\) of a \(0.0355 M\) solution of HCl is needed to reach the first equivalence point \((\mathrm{pH}=8.3)\) and another \(10.42 \mathrm{mL}\) is needed to reach the second equivalence point \((\mathrm{pH}=4.0) .\) If the alkalinity of the spring water is due only to the presence of carbonate and bicarbonate, what are the concentrations of each?

Step-by-Step Solution

Answer

Answer: The concentration of carbonate in the spring water is 0.00370 M, and the concentration of bicarbonate is 0.00091 M.

Step 1: Calculate moles of HCl at first equivalence point

At the first equivalence point, 2.56 mL of 0.0355 M HCl solution were added. We can calculate the moles of HCl at this point as follows: Moles of HCl = Volume (L) × Molarity Moles of HCl = (2.56 mL × (1 L / 1000 mL)) × 0.0355 mol/L = 0.00009108 mol

Step 2: Calculate moles of HCl at second equivalence point

At the second equivalence point, an additional 10.42 mL of HCl were added. We can calculate the total moles of HCl at this point as follows: Moles of HCl = Volume (L) × Molarity Moles of HCl = ((2.56 mL + 10.42 mL) × (1 L / 1000 mL)) × 0.0355 mol/L = 0.00046074 mol

Step 3: System of equations for carbonate and bicarbonate

At the first equivalence point, all bicarbonate ions have reacted with an equivalent amount of HCl to form carbonate ions. Therefore, moles of bicarbonate ions are equal to moles of HCl at the first equivalence point. Moles of bicarbonate = Moles of HCl at first equivalence point \(HCO_{3}^{-} = 0.00009108\) At the second equivalence point, all carbonate ions have reacted with an equivalent amount of HCl to become completely neutralized. Therefore, the moles of carbonate ions are equal to the difference between moles of HCl at the second and first equivalence points. Moles of carbonate = Moles of HCl at second equivalence point - Moles of HCl at first equivalence point \(CO_{3}^{2-} = 0.00046074 - 0.00009108 = 0.00036966\)

Step 4: Calculate concentrations of carbonate and bicarbonate

To find the concentrations of carbonate and bicarbonate ions, we can divide the moles of each ion by the volume of the water sample (100.0 mL). Concentration of bicarbonate (\([HCO_{3}^{-}]\)) = Moles of bicarbonate / Volume (L) \([HCO_{3}^{-}] = \frac{0.00009108}{100.0 \mathrm{mL} \times (1 \mathrm{L} / 1000 \mathrm{mL})} = 0.0009108 M\) Concentration of carbonate (\([CO_{3}^{2-}]\)) = Moles of carbonate / Volume (L) \([CO_{3}^{2-}] = \frac{0.00036966}{100.0 \mathrm{mL} \times (1 \mathrm{L} / 1000 \mathrm{mL})} = 0.0036966 M\) In conclusion, the concentration of carbonate in the spring water is \(0.00370 M\), and the concentration of bicarbonate is \(0.00091 M\).

Key Concepts

Equivalence Point

The equivalence point in a titration is a significant milestone. It indicates that the quantity of added titrant is stoichiometrically equal to the amount of substance being titrated in the solution. In simple terms, it's the point where the chemical reactions between the titrant and the solution are perfectly balanced.
For the alkalinity titration given, there are two equivalence points:

• The first equivalence point, reached at a pH of 8.3, indicates that all bicarbonate ions ( HCO_{3}^{-} ) have reacted with hydrochloric acid (HCl) to form carbonate ions ( CO_{3}^{2-} ).
• The second equivalence point, at a pH of 4.0, shows that all carbonate ions have reacted and fully neutralized, becoming part of carbon dioxide or another form.

Understanding these equivalence points is crucial for calculating the concentrations of carbonate and bicarbonate, as you can determine from the total amount of acid used in the process.

Carbonate Concentration

After reaching the first equivalence point in the titration, bicarbonate ions (HCO_{3}^{-}) are transformed into carbonate ions (CO_{3}^{2-}). However, to find the concentration of carbonate ions in solution after reaching the second equivalence point, the continuous addition of HCl is necessary. By that time, all carbonate ions are neutralized.
To calculate the carbonate concentration, use the additional moles of HCl added between the first and second equivalence points. This is because this amount of HCl represents the full neutralization of the carbonate ions. From the given solution, the moles of carbonate ions are:\[ CO_{3}^{2-} = 0.00046074 - 0.00009108 = 0.00036966 \] The concentration in the original 100 mL of spring water becomes:\[ [CO_{3}^{2-}] = \frac{0.00036966}{0.1} = 0.0036966 \, M \]

Bicarbonate Concentration

At the first equivalence point, all present bicarbonate ions have been transformed. Here, the moles of HCl equal the moles of bicarbonate ions in the hot spring water sample. Knowing the volume of the water sample allows you to determine their concentration.
In our problem, the moles of bicarbonate ions match the moles of HCl used to reach the first equivalence point:\[ HCO_{3}^{-} = 0.00009108 \, \text{mol} \]Once calculated, the concentration of bicarbonate in the sample is:\[ [HCO_{3}^{-}] = \frac{0.00009108}{0.1} = 0.0009108 \, M \] This calculation helps identify the initial alkalinity contribution from bicarbonate ions.

Moles of HCl

In chemical titration, calculating the moles of added titrant, here HCl, is vital. It helps reveal the concentrations of the sample's components, such as bicarbonate and carbonate. To determine these moles, the volume of HCl added at each equivalence point is multiplied by its molarity.
For the first equivalence point:

• Volume = 2.56 mL = 0.00256 L
• Moles of HCl = 0.00256 L \( \times 0.0355 \, \text{mol/L} \)\[ = 0.00009108 \, \text{mol} \]

For the second equivalence point:

• Total Volume = (2.56 + 10.42) mL = 0.01298 L
• Moles of HCl = 0.01298 L \( \times 0.0355 \, \text{mol/L} \)\[ = 0.00046074 \, \text{mol} \]

Understanding these calculations allows one to derive the concentrations of chemical species in the solution.