Problem 96 - Chemistry The Molecular Science

Question

You start with two \(1.00-\mathrm{L}\) samples: (1) pure water at \(\mathrm{pH}=7.00 ;\) and (2) lake water at \(\mathrm{pH}=6.96\) due to the presence of \(4.0 \times 10^{-5} \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}\) and \(1.6 \times 10^{-4} \mathrm{M}\) \(\mathrm{HCO}_{3}^{-}\). Then, \(10.0 \mathrm{~mL}\) of \(0.0010 \mathrm{M} \mathrm{HCl}\) was added to each sample. (a) Calculate the resulting \(\mathrm{pH}\) of the formerly pure water. (b) Calculate the resulting \(\mathrm{pH}\) of the lake water. (c) Which sample did not act as a buffer? Explain your answer.

Step-by-Step Solution

Answer

(a) 5.00; (b) 6.96; (c) Pure water did not act as a buffer.

Step 1: Calculate moles of HCl added

First, calculate the moles of HCl added to each sample. The concentration of HCl is 0.0010 M, and the volume added is 10.0 mL or 0.010 L. Use the formula for moles: \[ \text{moles of HCl} = 0.0010 \, \text{M} \times 0.010 \, \text{L} = 1.0 \times 10^{-5} \, \text{moles} \]

Step 2: Calculate pH change in pure water

The pure water initially has a pH of 7, meaning an initial \[ \text{\([H^+] = 10^{-7}\ M\)}. \] Add the moles of HCl to the pure water to get the new concentration of \([H^+]\): \[ \text{new } ext{[H^+]} = \frac{1.0 \times 10^{-5} \text{ moles} + 10^{-7} \text{ M} \times 1.00 \text{ L}}{1.010 \text{ L}} = \frac{1.0 \times 10^{-5} \text{ moles}}{1.010 \text{ L}} \approx 9.9\times 10^{-6}\ M \] Calculate the new pH: \[ \text{pH} = -\log_{10}(9.9 \times 10^{-6}) \approx 5.00 \]

Step 3: Calculate pH change in lake water

The lake water has a buffering capacity due to the presence of \( \text{H}_2\text{CO}_3 \) and \( \text{HCO}_3^{-} \). Use the Henderson-Hasselbalch equation \( \text{pH} = \text{pK}_a + \log\left(\frac{[\text{HCO}_3^{-}]}{[\text{H}_2\text{CO}_3]}\right) \) to calculate the pH after adding HCl. The \(\text{pK}_a\) for \(\text{H}_2\text{CO}_3\) is about 6.35.The addition of HCl converts some \(\text{HCO}_3^{-}\) into \(\text{H}_2\text{CO}_3\). Calculate the new concentrations:- Change in \([\text{HCO}_3^{-}]\): \(1.6 \times 10^{-4} - 1.0 \times 10^{-5} = 1.5 \times 10^{-4}\, \text{M}\)- Change in \([\text{H}_2\text{CO}_3]\): \(4.0 \times 10^{-5} + 1.0 \times 10^{-5} = 5.0 \times 10^{-5}\, \text{M}\)Substitute into the Henderson-Hasselbalch equation:\[\text{pH} = 6.35 + \log\left(\frac{1.5 \times 10^{-4}}{5.0 \times 10^{-5}}\right) \approx 6.96\]

Step 4: Compare buffering action

The pure water has a significant change in pH (from 7.00 to 5.00), while the lake water's pH remains nearly unchanged at approximately 6.96. The buffer system in the lake water effectively resists changes in pH on addition of HCl.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental formula used to predict the pH of a buffer solution. It establishes the relationship between the pH, pKa, and the concentrations of the acid and its conjugate base in solution.
To use this equation, one needs to know the pKa value of the acid and the ratio of the concentrations of the conjugate base and the acid. The equation is expressed as follows:
\[\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]
In this formula, \([\text{A}^-]\) represents the concentration of the conjugate base, while \([\text{HA}]\) stands for the concentration of the acid.
Applying this to the exercise, for the lake water, we calculated the pH after adding HCl. The carbonic acid (\(\text{H}_2\text{CO}_3\)) and bicarbonate (\(\text{HCO}_3^-\)) act as a buffer, allowing us to use the Henderson-Hasselbalch equation effectively. After finding the changes in concentrations caused by the interaction with HCl, substituting these values into the equation helps maintain the desired pH.

pH calculation

Calculating pH is crucial in many chemistry contexts, especially when examining acid-base reactions. pH is a measure of the hydrogen ion concentration and is calculated using the formula:\[\text{pH} = -\log_{10}([\text{H}^+])\]
In the original exercise, we calculated the resulting pH for both pure water and lake water after adding HCl. For pure water, HCl significantly increased the hydrogen ion concentration, changing the pH from 7.00 to around 5.00.
This change demonstrates how sensitive water is to added acids, as it has no buffering capacity. The pH indicates whether a solution is acidic or basic, which is vital knowledge when analyzing chemical reactions and environmental conditions.

acid-base equilibrium

Acid-base equilibrium plays a crucial role in maintaining stability within a chemical system. It involves the balance between acids and bases and how they neutralize each other when mixed.
In buffering systems, like the lake water in the example, both an acid and its conjugate base exist simultaneously. This equilibrium responds to the addition of an acid or base by shifting to restore pH balance. In this exercise, the presence of \(\text{H}_2\text{CO}_3\) and \(\text{HCO}_3^-\) creates a system that resists pH changes, exemplifying a buffer.
The equation for acid-base equilibrium can be generalized as:\[\text{HA} \rightleftharpoons \text{A}^- + \text{H}^+\]Here, \(\text{HA}\) represents the acid, and \(\text{A}^-\) is its conjugate base. The reversible nature of reactions depicted in the equilibrium helps in maintaining consistent pH levels when acids or bases are added, as seen with the lake water.