Question

Working with Buffers A buffer contains \(0.010\) mol of lactic acid \(\left(\mathrm{p} K_{\mathrm{n}}=3.56\right)\) and \(0.050\) mol of sodium lactate per liter. a. Calculate the pH of the buffer. b. Calculate the change in \(\mathrm{pIl}\) after adding \(5.0 \mathrm{~mL}\) of \(0.5\) M MCl to 1 L of the huffer. c. What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?

Short answer

a. 4.26 b. 0.12 pH change c. 4.40 pH change in pure water

Step-by-step solution

Understanding the Buffer Solution

Buffers consist of a weak acid and its conjugate base. In this problem, lactic acid (HC₃H₅O₃) is the weak acid and sodium lactate (C₃H₅O₃⁻Na⁺) is its conjugate base. The Henderson-Hasselbalch equation will be used to calculate the pH of this buffer.

Using the Henderson-Hasselbalch Equation for pH Calculation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \] In this buffer: -\([\text{Acid}] = 0.010 \; \text{mol/L}\) -\([\text{Base}] = 0.050 \; \text{mol/L}\) Thus:\[ \text{pH} = 3.56 + \log \left( \frac{0.050}{0.010} \right) \]\[ \text{pH} = 3.56 + \log (5) \] \[ \text{pH} = 3.56 + 0.70 = 4.26 \]

Determining pH Change after HCl Addition

When 5.0 mL of 0.5 M HCl is added, moles of HCl = 0.0025 mol. Adding HCl will react with sodium lactate:\[ \text{HCl} + \text{C}_3\text{H}_5\text{O}_3^- \rightarrow \text{HC}_3\text{H}_5\text{O}_3 + \text{Cl}^- \] Post-reaction:- Initial moles of conjugate base: 0.050 mol- Reacted moles: 0.0025 mol- Remaining moles of conjugate base: 0.050 - 0.0025 = 0.0475 mol- Initial moles of acid: 0.010 mol- Moles formed: 0.0025 mol- Total moles of acid: 0.010 + 0.0025 = 0.0125 molNow apply Henderson-Hasselbalch equation:\[ \text{pH} = 3.56 + \log\left(\frac{0.0475}{0.0125}\right)\]\[ \text{pH} = 3.56 + \log(3.8) \]\[ \text{pH} = 3.56 + 0.58 = 4.14 \]Therefore, the pH change is \(4.26 - 4.14 = 0.12\).

Calculating pH change in Pure Water

Adding 0.0025 mol of HCl to pure water creates a solution with: \[ \text{Concentration of H}^+ = \frac{0.0025\, \text{mol}}{1.005\, \text{L}} = 0.0024876\, \text{M} \]The pH of pure water after adding HCl is:\[ \text{pH} = -\log(0.0024876) \approx 2.60 \]. The pH change from 7.00 (neutral water) is approximately 4.40.

Key concepts

Buffer Solutions

Buffers are fascinating solutions that can resist changes in pH when small amounts of an acid or a base are added. This characteristic is crucial in many biological and chemical systems, as maintaining a consistent pH is often necessary for proper function. Buffers typically consist of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid.

They work by neutralizing added acids or bases. For example, if an acid is added to the buffer, the conjugate base present in the buffer will neutralize it. Conversely, if a base is added, the weak acid in the buffer will neutralize the base. This neutralization process helps maintain the stable pH of the solution.

In the example provided in the exercise, the buffer solution is composed of lactic acid (the weak acid) and sodium lactate (the conjugate base). This specific combination allows the solution to maintain a relatively stable pH even when an external acid such as HCl is introduced.

Weak Acids

Weak acids, like lactic acid in our example, are not as efficient at donating protons ( H⁺ ) as strong acids. They partially dissociate in water, meaning only a fraction of the acid molecules release H⁺ ions.

This partial dissociation is what makes weak acids important in buffer solutions. The ability to donate and accept H⁺ ions in small amounts is key to maintaining equilibrium and ensuring the buffer's effectiveness. Unlike strong acids, weak acids are better suited for buffers because they don't completely overwhelm the system with protons.

A weak acid's strength, and consequently its ability to donate H⁺ ions, is often measured using the ext{p}K_a value. In our exercise, lactic acid has a ext{p}K_a of 3.56, signifying how it balances between the dissociated and undissociated states in the solution.

pH Calculation

Calculating the pH of buffer solutions involves understanding the ratio of the concentrations of the conjugate base and the weak acid. This is where the Henderson-Hasselbalch equation comes into play: \[ \text{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]

This equation is a powerful tool for finding the pH of a buffer system because it takes into account both the inherent ext{p}K_a of the acid and the ratio of the concentrations of the conjugate base and the weak acid. By inserting the values from the exercise where [ ext{Base}] = 0.050 ext{ mol/L} and [ ext{Acid}] = 0.010 ext{ mol/L}, and noting that ext{p}K_a is 3.56 for lactic acid, we can calculate the pH of the buffer to be approximately 4.26.

Whenever there is a change in the concentration of either the weak acid or the conjugate base, like when HCl is added, we recalculate using the same equation to understand the new pH, showcasing the buffer's adjusting capacity.

Conjugate Acids and Bases

Conjugate acids and bases play a crucial role in the chemistry of buffer solutions. Whenever an acid donates a proton ( H⁺ ), its product is called a conjugate base. Similarly, when a base accepts a proton, its product is a conjugate acid.

In our buffer system involving lactic acid and sodium lactate, lactic acid ( HC₃H₅O₃ ) acts as the weak acid. When it donates a proton, it forms its conjugate base, lactate ion ( C₃H₅O₃^- ). The sodium lactate in the buffer provides these lactate ions.

This relationship forms a balance: the weak acid and its conjugate base work in tandem to resist drastic changes in pH. For example, when HCl is added to the buffer, the lactic acid increases slightly while the conjugate base decreases, but the overall pH remains fairly steady. This dynamic interplay is the hallmark of a well-functioning buffer system.