Problem 90 - Chemistry and Chemical Reactivity

Question

What volume of 0.200 M HCl must be added to \(500.0 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{NH}_{3}\) to have a buffer with a pH of \(9.00 ?\)

Step-by-Step Solution

Answer

400.0 mL of 0.200 M HCl must be added.

Step 1: Calculate the Ratio Using the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by \[pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]For \(\mathrm{NH}_3\) and \(\mathrm{NH}_4^+\), the \(pK_a = 9.25\). We set up the equation with \(pH = 9.00\):\[9.00 = 9.25 + \log \left( \frac{[\mathrm{NH}_3]}{[\mathrm{NH_4^+}]} \right)\]Rearrange to find the log ratio:\[\log \left( \frac{[\mathrm{NH}_3]}{[\mathrm{NH_4^+}]} \right) = 9.00 - 9.25 = -0.25\]

Step 2: Convert the Logarithmic Ratio to an Exponential Ratio

Convert the logarithmic equation to its exponential form:\[\frac{[\mathrm{NH}_3]}{[\mathrm{NH_4^+}]} = 10^{-0.25} \approx 0.562\]This means that for each mole of \(\mathrm{NH}_3\), there are approximately \(0.562\) moles of \(\mathrm{NH_4^+}\).

Step 3: Determine Moles of NH3 and HCl Required

Initially, we have \(0.250 \times 0.500 = 0.125\) moles of \(\mathrm{NH}_3\). Let \(x\) be the moles of \(\mathrm{NH_4^+}\) formed by adding \(\mathrm{HCl}\).\[\frac{0.125 - x}{x} = 0.562\]

Step 4: Solve the Equation for x

Solve for \(x\):\[0.125 - x = 0.562x\]\[0.125 = 1.562x\]\[x \approx \frac{0.125}{1.562} \]\[x \approx 0.0800 \text{ moles}\]This is the amount of \(\mathrm{NH_4^+}\) (and \(\mathrm{HCl}\) added).

Step 5: Calculate the Volume of HCl Needed

Since \(x = 0.0800\) moles of \(\mathrm{HCl}\) are needed and its concentration is \(0.200 \mathrm{M}\):\[\text{Volume} = \frac{0.0800}{0.200} = 0.400 \text{ L}\]Convert this to milliliters: \(0.400 \times 1000 = 400.0 \text{ mL}\).

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a key tool in chemistry for calculating the pH of a buffer solution. It relates the pH, pKa, and the concentrations of an acid and its conjugate base in solution. The equation is expressed as \[pH = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right)\]This equation is highly valuable for understanding how the ratio of the base to acid affects the pH of a solution. By applying it, one can easily determine changes in pH in response to variations in the concentrations of the buffer components.

• It gives a direct relationship between pH and the concentration ratio of the buffer components.
• It helps predict how changes in concentration affect pH without having to conduct experiments.
• Knowing the pKa value is essential, as it is a measure of the strength of the acid component in the buffer.

Understanding the Henderson-Hasselbalch equation is crucial for solving problems involving buffer solutions and can help in various practical applications like making buffer solutions of desired pH.

pH Calculation

Calculating the pH of a solution involves understanding the concentration of hydrogen ions in the solution. In buffer solutions, this calculation is often simplified using the Henderson-Hasselbalch equation. When given the pKa and the concentrations of an acid and its base form, you can quickly determine the pH. For example, if we know the concentrations of ammonium ion (\(\text{NH}_4^+\)) and ammonia (\(\text{NH}_3\)), using the equation:

• We plug in known values of pH, which is set as a goal (9.00 in this example).
• Calculate the log ratio and rearrange the equation to find the balance between acid and base concentrations required.
• Translate the logarithmic ratio into an exponential, allowing calculation of actual concentrations.

This calculation provides insight into how much of each component is required to maintain the desired pH. It is especially useful in biochemical processes where maintaining a stable pH is crucial.

Acid-Base Equilibrium

The concept of acid-base equilibrium is at the core of understanding buffer solutions. It involves the balance between weak acids and their conjugate bases in solution. Buffers resist changes in pH when small amounts of acid or base are added because they contain both an acidic component and a basic component that can neutralize additions of either acid or base. In the context of the given exercise:

• Ammonia (\(\text{NH}_3\)) acts as a base, while its conjugate acid form, ammonium (\(\text{NH}_4^+\)), regulates added acids.
• The objective was to reach a specific pH (9.00), maintaining equilibrium via the proper balance of \(\text{NH}_3\) and \(\text{NH}_4^+\).

Proper understanding of acid-base equilibria is essential for predicting how solutions will react upon adding different chemicals or changes in conditions, ensuring stability in chemical reactions and formulations.