Problem 85 - Chemistry: The Central Science

Question

If \(40.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is titrated with \(0.100 \mathrm{M}\) \(\mathrm{HCl}\), calculate (a) the \(\mathrm{pH}\) at the start of the titration; (b) the volume of \(\mathrm{HCl}\) required to reach the first equivalence point and the predominant species present at this point; (c) the volume of \(\mathrm{HCl}\) required to reach the second equivalence point and the predominant species present at this point; (d) the \(\mathrm{pH}\) at the second equivalence point.

Step-by-Step Solution

Answer

(a) The initial pH is 13.3010. (b) The volume of HCl required to reach the first equivalence point is 0.0400 L, and the predominant species is HCO₃⁻. (c) The volume of HCl required to reach the second equivalence point is 0.0800 L, and the predominant species is H₂CO₃. (d) The pH at the second equivalence point is 3.53.

Step 1: Find pH at the start of the titration

For this, we need to know the concentration of hydroxide ions [OH⁻]. Since Na₂CO₃ is a strong base and completely dissociates in water, its concentration can be found using the formula: Concentration of Na₂CO₃ (M) = 0.100M This will dissociate to form 2 moles of sodium ion (Na⁺) and 1 mole of carbonate ion (CO₃²⁻). Considering that water has a Kw = 1.0 x 10⁻¹⁴, we can then find the pOH: pOH = -log[OH⁻] Initial pOH = -log(2 x 0.100) = -log(0.200) pOH = 0.6990 Now we can find the pH using the formula: pH + pOH = 14 Initial pH = 14 - pOH = 14 - 0.6990 = 13.3010 (a) The pH at the start of the titration is 13.3010.

Step 2: Calculate the volume of HCl required to reach the first equivalence point

At the first equivalence point, moles of base must be equal to moles of acid. Using the formula: Moles = Molarity x Volume Moles of Na₂CO₃ = 0.100M x 0.0400L = 0.00400 moles For the first equivalence point, the mole ratio of Na₂CO₃ to HCl is 1:1. Therefore, Moles of HCl = 0.00400 moles Now we can calculate the volume of HCl needed: Volume of HCl = Moles / Molarity = 0.00400 moles / 0.100M = 0.0400 L (b) The volume of HCl required to reach the first equivalence point is 0.0400 L. At this point, the predominant species is HCO₃⁻ (hydrogen carbonate ion).

Step 3: Calculate the volume of HCl required to reach the second equivalence point

At the second equivalence point, there will be twice the moles of hydrochloric acid used compared to moles of sodium carbonate: Moles of HCl = 2 x Moles of Na₂CO₃ = 2 x 0.00400 moles = 0.00800 moles Now we can calculate the volume of HCl needed: Volume of HCl = Moles / Molarity = 0.00800 moles / 0.100M = 0.0800 L (c) The volume of HCl required to reach the second equivalence point is 0.0800 L. At this point, the predominant species is H₂CO₃ (carbonic acid).

Step 4: Calculate the pH at the second equivalence point

To find the pH at the second equivalence point, we need to find the concentration of H⁺ ions. At this point, the H₂CO₃ will dissociate to form H⁺ ions and HCO₃⁻ ions. Considering the equilibrium constant Ka₁ of H₂CO₃ is 4.45 x 10⁻⁷: Ka₁ = [H⁺][HCO₃⁻] / [H₂CO₃] As we have the same number of moles of H₂CO₃ and HCO₃⁻, we can write: Ka₁ = (2x)(2x) / (0.00400-x) 4.45 x 10⁻⁷ = 4x² / (0.00400-x) Solving the quadratic equation, we find x = 2.97 x 10⁻⁴ Now, we can find the pH: pH = -log[H⁺] = -log(2.97 x 10⁻⁴) = 3.53 (d) The pH at the second equivalence point is 3.53.

Key Concepts

pH calculation

In the beginning of an acid-base titration involving sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and hydrochloric acid (HCl), determining the pH is crucial. Sodium carbonate is a strong base that completely dissociates in water, forming sodium ions (\(\text{Na}^+\)) and carbonate ions (\(\text{CO}_3^{2-}\)). The concentration of hydroxide ions (\([\text{OH}^-]\)), is derived from the carbonate ion. By calculating \(\text{pOH}\) using the formula:

• \(\text{pOH} = -\log[\text{OH}^-]\)

We find the \(\text{pOH}\) and then use the relationship:

• \(\text{pH} + \text{pOH} = 14\)

This helps us to derive the initial pH of the solution. Knowing how to calculate pH ensures you're understanding how solutions change during titration.

equivalence point

In a titration, an equivalence point is where the amount of titrant added exactly reacts with the substance in the solution. For the first equivalence point in this exercise, one molecule of sodium carbonate reacts with one molecule of HCl, resulting in the formation of hydrogen carbonate ions (\(\text{HCO}_3^-\)). This can be represented by the reaction:

• \(\text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl}\)

In the second equivalence point, this hydrogen carbonate ion reacts again with HCl to form carbonic acid (\(\text{H}_2\text{CO}_3\)).

• \(\text{NaHCO}_3 + \text{HCl} \rightarrow \text{H}_2\text{CO}_3 + \text{NaCl}\)

At equivalence points, the solution composition changes significantly, which is important for determining titration volumes and calculating resultant pH.

carbonate ion

The carbonate ion (\(\text{CO}_3^{2-}\)) is a part of the reaction process between sodium carbonate and HCl in titration. It serves as the starting point in this acid-base titration setup. As titration progresses:

• It initially reacts with HCl, forming hydrogen carbonate (\(\text{HCO}_3^-\)).
• As more acid is added, \(\text{HCO}_3^-\) can further react to produce carbonic acid (\(\text{H}_2\text{CO}_3\)).

Understanding the role of the carbonate ion allows insight into the progression of the titration and prediction of resulting pH at different stages. The carbonate ion acts as a buffer, influencing amounts of HCl needed and the species formed at equivalence points.

titration curve

A titration curve is a graphical representation of pH change throughout the course of a titration. For a sodium carbonate and hydrochloric acid titration, two distinct regions signify two equivalence points:

• The curve starts at a high pH, reflecting the basic nature of the solution.
• At the first equivalence point, a significant drop in pH occurs as \(\text{Na}_2\text{CO}_3\) is converted to \(\text{HCO}_3^-\).
• As titration progresses to the second equivalence point, pH decreases further due to formation of \(\text{H}_2\text{CO}_3\).

Titration curves are invaluable for visualizing changes in pH and identifying points of reactions completing. They help portray how and when reactions reach completion, and the stability of the solution.