Calculate the \(\mathrm{pH}\) of a solution of given mixtures; (a) \(\left(4 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}+6 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COONa}\right)\) in \(100 \mathrm{~mL}\) of mixture; \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}\) (b) \(5 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MOH}+250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MCl}\); \(K_{\mathrm{a}}\) for \(M \mathrm{OH}=1.8 \times 10^{-5}\) (c) \(\left(0.25\right.\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}+0.35\) mole of \(\mathrm{CH}_{3} \mathrm{COONa}\) ) in \(500 \mathrm{ml}\) mixture; \(K_{\text {a }}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=3.6 \times 10^{-4}\)

Short Answer

Expert verified
Mixture (a) pH can be found using the Henderson-Hasselbalch equation after converting grams to moles and finding concentration. Mixture (b) will react to neutralization if MOH and MCl are in equal molar amounts; if not, the excess will determine the pH. Mixture (c) also uses Henderson-Hasselbalch after establishing the acid and salt concentration.

Step by step solution

01

- Convert mass to moles for mixture (a)

Calculate the moles of each component in the mixture using the formula mole = mass / molar mass. The molar masses are 60.05 g/mol for acetic acid (CH_3COOH) and 82.03 g/mol for sodium acetate (CH_3COONa). For CH_3COOH, mole = 4 g / 60.05 g/mol and for CH_3COONa mole = 6 g / 82.03 g/mol.
02

- Apply Henderson-Hasselbalch equation for mixture (a)

Use the Henderson-Hasselbalch equation to find the pH: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. The pKa = -log(Ka) and the volumes of the acid and salt solutions are the same.
03

- Solve for pH for mixture (a)

Calculate the pH using the concentrations of CH_3COOH and CH_3COONa. Convert the moles to concentrations by dividing by the volume of the solution (in liters), then plug the values into the Henderson-Hasselbalch equation.
04

- Apply the concept of dilution for mixture (b)

Determine the final concentrations of MOH and MCl after mixing by using the dilution formula M1V1 = M2V2. Calculate the new volumes and then obtain the new concentrations.
05

- Calculate pH for strong acid and base reaction in mixture (b)

The reaction between MOH and MCl will produce water and M2(NO3)2. Since they are strong acid and base, the reaction goes to completion. If there's excess of OH- or H+, calculate the pH based on the excess moles.
06

- Mixing of acid and salt for mixture (c)

Find the amount of acetic acid that has reacted by subtracting the smaller number of moles (from the limiting reagent) from the larger one. This will give the number of moles of unreacted acetic acid and the produced salt.
07

- Calculate concentrations for mixture (c)

Divide the moles of unreacted acetic acid and the formed salt by the volume of the solution in liters to get their concentrations.
08

- Solve for pH for mixture (c)

Using the Henderson-Hasselbalch equation, calculate the pH of the solution for mixture (c) using the concentrations of the acid and salt obtained in the previous step.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is vital when attempting to calculate the pH of buffer solutions. This equation provides a direct connection between the pH of a solution and the ratio of the concentration of the salt (conjugate base) to the concentration of the acid. It's expressed as \[\mathrm{pH} = \mathrm{pKa} + \log(\frac{[A^-]}{[HA]})\] where \(\mathrm{pKa}\) is the acid dissociation constant, \([A^-]\) represents the concentration of the conjugate base, and \([HA]\) is the concentration of the acid.

To use this equation effectively in pH calculation problems, you must first understand what \(\mathrm{pKa}\) stands for and how to calculate it. The \(\mathrm{pKa}\) is the negative logarithm of the acid dissociation constant \(K_a\), which indicates the strength of an acid. Typically, a lower \(\mathrm{pKa}\) value indicates a stronger acid. When applying the Henderson-Hasselbalch equation to mixture (a) in the given exercise, we first converted the mass of acetic acid and sodium acetate to moles, followed by deriving their concentrations in the solution. With these values, we could easily find the pH.
Acid-Base Equilibria
The concept of acid-base equilibria is at the heart of understanding chemical reactions involving acids and bases, and consequently how to calculate pH. Acids are substances that release hydrogen ions (\(H^+\)) in solution, while bases provide hydroxide ions (\(OH^-\)). When an acid reacts with a base, they form a salt and water in a neutralization reaction.

In mixture (b) of our exercise, we dealt with such an acid-base reaction. A strong base \(MOH\) reacted completely with a strong acid \(MCl\), resulting in water and a salt. The pH of the resulting solution depends on whether an excess of acid or base is present after the reaction. Understanding the stoichiometry of the reaction helps in calculating which reactant is in excess and by how much, thus enabling us to calculate the pH of the final mixture accurately.
Molarity and Dilution
Molarity, often represented by M, is a measure of concentration in chemistry, indicating the number of moles of solute per liter of solution. The concept of molarity is crucial when we prepare solutions and perform dilutions, as it allows us to calculate the concentration changes that occur when solutions of different volumes and concentrations are mixed together.

The key formula for dilutions is \(M_1V_1 = M_2V_2\), where \(M_1\) and \(V_1\) are the molarity and volume of the initial solution, and \(M_2\) and \(V_2\) are the molarity and volume after dilution. In part (b) of our exercise, we used this concept to adjust the concentrations of the strong base and acid after mixing. Knowing the final molarity of the reactants allowed us to understand the remaining quantities post-reaction and solve for the pH effectively, showcasing how intertwined these concepts are in practical applications of chemistry. Understanding how to manipulate molarity through dilutions is an essential skill for tackling pH calculation problems.

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