Question

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to produce a solution buffered at each pH. a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) b. \(\mathrm{pH}=4.20\) c. \(\mathrm{pH}=5.00\)

Short answer

To create buffered solutions at the given pH values, the following amount of moles of HCl(g) must be added to \(1.0 \mathrm{~L}\) of \(1.0 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\): a) For pH \(= 4.74\), add 1 mole of HCl. b) For pH \(= 4.20\), add approximately 0.223 moles of HCl. c) For pH \(= 5.00\), add approximately 0.645 moles of HCl.

Step-by-step solution

Find the \(K_{a}\) and \(\mathrm{p}K_{\mathrm{a}}\) of acetic acid

The \(K_{a}\) and \(\mathrm{p}K_{\mathrm{a}}\) of acetic acid are given as: \[K_{a} = 1.8 \times 10^{-5}\] \[\mathrm{p}K_{\mathrm{a}} = -\log K_{\mathrm{a}} = -\log(1.8 \times 10^{-5}) = 4.74\]

Use the Henderson-Hasselbalch equation to find the ratio

The Henderson-Hasselbalch equation is given by: \[{\mathrm{pH}}={\mathrm{p}K_{\mathrm{a}}} + \log\left(\frac{\left[\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\right]}\right)\] Rearrange the equation to find the ratio: \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right] = 10^{\mathrm{(pH - pK}_{a}\mathrm{)}\mathrm{)}}\] Now, we will use this equation to find the ratio for each given pH value. a. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} = 4.74\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{a} = 10^{(4.74-4.74)} = 10^0 = 1\] b. \(\mathrm{pH}=4.20\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{b} = 10^{(4.20-4.74)} = 10^{-0.54} \approx 0.287\] c. \(\mathrm{pH}=5.00\) \[\left[\frac{\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}}{\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}}\right]_{c} = 10^{(5.00-4.74)} = 10^{0.26} \approx 1.819\]

Find the moles of \(\mathrm{HCl}\) needed

Since we have a \(1.0 \mathrm{~L}\) solution of \(1.0 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), initially, we have 1 mole of \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) and 0 moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\). a) For pH \(= 4.74\), we need an equal amount of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), which is 1 mole. Adding 1 mole of \(\mathrm{HCl}\) to 1 mole of acetate ions gives us 1 mole of acetic acid, as they react with a 1:1 ratio. b) For pH \(= 4.20\), we need a \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) to \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) ratio of 0.287. Let's use x to represent the moles of \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[\frac{x}{1.0-x} = 0.287\] \[x = 0.287(1.0-x)\] \[x = 0.287 - 0.287x\] \[1.287x = 0.287\] \[x \approx 0.223\] So, we need to add approximately 0.223 moles of \(\mathrm{HCl}\). c) For pH \(= 5.00\), we need a \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) to \(\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}\) ratio of 1.819. \[\frac{x}{1.0-x} = 1.819\] \[x = 1.819(1.0-x)\] \[x = 1.819 - 1.819x\] \[2.819x = 1.819\] \[x \approx 0.645\] So, we need to add approximately 0.645 moles of \(\mathrm{HCl}\). In conclusion, we need to add: - For pH \(= 4.74\), 1 mole of \(\mathrm{HCl}\). - For pH \(= 4.20\), approximately 0.223 moles of \(\mathrm{HCl}\). - For pH \(= 5.00\), approximately 0.645 moles of \(\mathrm{HCl}\).

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is like learning a secret language that reveals the balance point of acids and bases in a solution. In essence, this formula serves as a pH calculator for buffered solutions, which are solutions resistant to pH change upon the addition of small amounts of acid or base.

The equation is written as: \[\begin{equation}\mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right),\end{equation}\]where \( \mathrm{pH} \) is the measure of the acidity or basicity of the solution, \( \mathrm{p}K_a \) is the negative logarithm of the acid dissociation constant (\( K_a \)), \( [\mathrm{A}^-] \) is the concentration of the conjugate base, and \( [\mathrm{HA}] \) is the concentration of the acid.

It's like putting two kids on a seesaw. The \( \mathrm{p}K_a \) represents the fulcrum's position, while the concentrations of the acid and conjugate base are the weights of the kids. When they're balanced, you get the buffered pH. If you adjust the weights (concentrations), the seesaw tips, and the pH changes in accordance.

Moles Calculation

Diving into the world of moles calculation is a must for any budding chemist or anyone curious about the quantities involved in chemical reactions. The mole is a standard unit in chemistry that helps us count atoms, molecules, or other chemical units by weighing them. It's like having a giant container that always holds the same number of marbles, no matter what.To calculate moles, you can use the simple formula:\[\begin{equation}\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}.\end{equation}\]Or if you're dealing with solutions, the formula is:\[\begin{equation}\text{moles} = \text{concentration (M)} \times \text{volume (L)}.\end{equation}\]In our exercise, we needed to add certain moles of \( \mathrm{HCl} \) to adjust the pH. By deducing the necessary ratio of acid to base, we can determine the existing amount of conjugate base and then calculate how much acid (in moles) we need to add to reach the desired pH.

Acid-Base Equilibrium

When exploring the acid-base equilibrium, we venture into a tug-of-war between what a particular acid is willing to give up—its hydrogen ions (H+)—and how a base is ready to accept them. This equilibrium is a delicate balance of give and take, maintained by reversible chemical reactions that occur in the solution.

Imagine you have a number of dancers (acid molecules) and a number of invitations (hydrogen ions). When the music starts (the solution is mixed), some dancers hand out their invitations (the acid dissociates). At the same time, other dancers (the conjugate bases) who got tired may want to return (recombine with hydrogen ions to form the acid).This ongoing dance is described by the acid dissociation constant (\( K_a \)), which tells us how strong an acid is, or essentially, how eager the dancers are to keep dancing or take a rest. The larger the value of \( K_a \), the stronger the acid and the more the equilibrium lies towards the conjugate base. When it comes to calculations, knowing the \( K_a \) and the volumes and concentrations of the acids and bases involved allows us to predict the pH of the solution and manage the equilibrium to suit our needs.