Problem 36 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

What is the \(\mathrm{pH}\) of \(4 \times 10^{-3} \mathrm{M}-\mathrm{Y}(\mathrm{OH})_{2}\) solution assuming the first dissociation to be \(100 \%\) and second dissociation to be \(50 \%\), where \(Y\) represents a metal cation? \((\log 2=0.3, \log 3=0.48)\) (a) \(11.78\) (b) \(11.22\) (c) \(2.22\) (d) \(2.78\)

Step-by-Step Solution

Answer

The pH of the solution is 11.78.

Step 1: Understanding the problem

We are given a solution of a metal hydroxide, \(Y(OH)_2\), which is a dibasic base. This means it can undergo two steps of dissociation. In this problem, the first dissociation is given as complete (100%), while the second dissociation is 50%. We need to calculate the \(pH\), given the initial concentration of \(Y(OH)_2\) and the dissociation percentages. To find the \(pH\), we need to determine the concentration of \(OH^-\) ions in the solution.

Step 2: Calculating the concentration of \(OH^-\) ions after first dissociation

Given that the first dissociation is 100%, the concentration of \(OH^-\) ions from the first dissociation is equal to the initial concentration of \(Y(OH)_2\). For every mole of \(Y(OH)_2\) that dissociates, one mole of \(OH^-\) is produced. Therefore, the concentration of \(OH^-\) ions after the first dissociation is \(4 \times 10^{-3} \mathrm{M}\).

Step 3: Calculating the concentration of \(OH^-\) ions after second dissociation

For the second dissociation, which is 50%, half of the \(Y(OH)^+\) (formed from the first dissociation) will dissociate to produce additional \(OH^-\) ions. The concentration of \(Y(OH)^+\) is equal to the concentration of \(OH^-\) ions from the first dissociation. Thus, the additional \(OH^-\) ions produced is half of \(4 \times 10^{-3} \mathrm{M}\), which is \(2 \times 10^{-3} \mathrm{M}\).

Step 4: Calculating the total concentration of \(OH^-\) ions

The total concentration of \(OH^-\) ions will be the sum of \(OH^-\) ions from the first and the second dissociation step. So, the total concentration of \(OH^-\) ions is \((4 \times 10^{-3} \mathrm{M} + 2 \times 10^{-3} \mathrm{M}) = 6 \times 10^{-3} \mathrm{M}\).

Step 5: Calculating the \(pOH\) of the solution

The \(pOH\) is given by \(pOH = -\log [OH^-]\). Calculating the \(pOH\) for the total concentration of \(OH^-\) ions, we have \(pOH = -\log (6 \times 10^{-3})\).

Step 6: Finding the \(pOH\) using logarithms

We can separate the logarithm into two parts: \(\log 6 + \log 10^{-3}\). Using the given logarithm values: \(-\log 10^{-3} = -(-3) = 3\) and \(-\log 6 = -\log (2\times3) = -\log 2 - \log 3 = -(0.3 + 0.48) = -0.78\), we can determine that \(pOH = 3 - 0.78 = 2.22\).

Step 7: Calculating the \(pH\) from the \(pOH\)

The relationship between \(pH\) and \(pOH\) is given by the equation \(pH + pOH = 14\). Substituting the calculated \(pOH\) into this equation gives us \(pH = 14 - 2.22 = 11.78\).

Key Concepts

Dissociation of Weak Bases

In any aqueous solution, understanding the dissociation of weak bases is important for predicting the behavior of the solution. A weak base does not completely ionize in water, instead it establishes an equilibrium between the un-ionized base and the ions produced.

Upon adding a weak base to water, it reacts with water molecules to form hydroxide ions (\(OH^-\)) and the conjugate acid of the base. The degree to which this reaction occurs is measured by the base dissociation constant (\(K_b\)). In the given exercise, the base (\(Y(OH)_2\)) is a weak base that undergoes two dissociations. The first dissociation is said to be complete, which is unusual for weak bases, while the second dissociation is 50% complete, demonstrating its weaker tendency to dissociate.

Bases with multiple hydroxyl groups, like the dibasic base in our problem, undergo stepwise dissociation. The first hydroxyl group dissociates more readily than the second, as reflected in the differing percentages of dissociation presented in the problem. This stepwise dissociation significantly affects the hydroxide ion concentration in solution and thus, the resulting pH.

pOH to pH Conversion

Understanding the inverse relationship between pOH and pH is fundamental to solving many chemistry problems. The scale of pOH measures the basicity of a solution, with lower pOH values indicating stronger basicity. pH, on the other hand, measures acidity, with lower pH values indicating higher acidity.

The exercise showed us how to calculate pOH first and then use it to find the pH, which was achieved through the relationship that pH + pOH = 14. This relationship is a consequence of the water dissociation constant (\(K_w\)), which is the product of the concentrations of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)) in water at a given temperature, typically 25°C.

By converting pOH to pH, we're essentially shifting our perspective from the basicity to the acidity of the solution. This conversion is useful in several applications, including titrations and buffer solutions, making it a vital tool for students to master in the realm of chemistry.