Problem 115 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

How does the electrical conductivity of \(20 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{MgSO}_{4}\) change when \(0.5 \mathrm{M}-\mathrm{Ba}(\mathrm{OH})_{2}\) solution is gradually added in it, to excess? (a) decreases continuously (b) increases continuously (c) increases and then decreases (d) decreases am

Step-by-Step Solution

Answer

The conductivity first increases due to the increase in \( \mathrm{OH}^{-} \) ions and then decreases as \( \mathrm{BaSO}_{4} \) and \( \mathrm{Mg(OH)}_{2} \) precipitates form, thus the correct answer is (c) increases and then decreases.

Step 1 - Understanding the reaction

Recognize that magnesium sulfate, \( \mathrm{MgSO}_{4} \), is a strong electrolyte and dissociates completely into \( \mathrm{Mg}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \). Barium hydroxide, \( \mathrm{Ba(OH)}_{2} \), is also a strong electrolyte and dissociates into \( \mathrm{Ba}^{2+} \) and \(2 \mathrm{OH}^{-} \). When \( \mathrm{Ba(OH)}_{2} \) is added to the \( \mathrm{MgSO}_{4} \) solution, a precipitation reaction occurs, forming \( \mathrm{BaSO}_{4} \) and \( \mathrm{Mg(OH)}_{2} \).

Step 2 - Predicting conductivity changes

Initially, as \( \mathrm{Ba(OH)}_{2} \) is added, \( \mathrm{OH}^{-} \) ions increase, which can increase the conductivity. However, as the reaction continues, \( \mathrm{BaSO}_{4} \) and \( \mathrm{Mg(OH)}_{2} \), both being insoluble, precipitate out of solution, thereby reducing the number of free ions that contribute to conductivity. Thus, after initially increasing due to the added \( \mathrm{OH}^{-} \), conductivity will then decrease as the precipitation reaction proceeds.

Step 3 - Selecting the correct answer

Based on the analysis, the conductivity will first increase because of the addition of \( \mathrm{OH}^{-} \) ions from \( \mathrm{Ba(OH)}_{2} \). After reaching a maximum, the conductivity will start to decrease as the insolouble precipitates, \( \mathrm{BaSO}_{4} \) and \( \mathrm{Mg(OH)}_{2} \), form and remove ions from the solution. Therefore, the correct answer is that the conductivity increases and then decreases, corresponding to option (c).

Key Concepts

Precipitation Reactions

In the fascinating world of chemistry, precipitation reactions belong to the broader category of double displacement reactions. These occur when two ionic compounds in solution are mixed together resulting in the formation of at least one insoluble product, known as the precipitate. This is notable for causing a visible change in the solution, often characterized by the appearance of a solid from a previously clear liquid.

For instance, when a solution of barium hydroxide is added to a solution of magnesium sulfate, both solutions being composed of strong electrolytes, a solid precipitate of barium sulfate forms. During this process, magnesium hydroxide also forms as a precipitate, although it may not immediately drop out of solution due to its slightly higher solubility.

To accurately predict the products of a precipitation reaction, it is essential to know the solubility rules. Certain compounds, such as those containing nitrate (NO_3^-) ion or alkali metal cations like sodium (Na^+) and potassium (K^+), are typically soluble, while others, such as carbonates (CO_3^{2-}) and sulfates (SO_4^{2-}), might be insoluble, depending on what they are combined with.

Conductivity of Electrolyte Solutions

Diving deeper into electrolyte solutions, they are fundamental in conducting electricity due to the presence of free-moving ions within them. The measure of a solution’s ability to conduct electricity is referred to as electrical conductivity.

When discussing electrolyte solutions, you might come across terms like 'weak' or 'strong' electrolytes. These labels identify how completely ions dissociate in solution. Strong electrolytes, like NaCl and HCl, split into cations and anions entirely, which contributes to a higher conductivity. Meanwhile, weak electrolytes do not fully dissociate, resulting in lower conductivity.

In the classroom scenario, as you add more barium hydroxide to the magnesium sulfate, the increase in free OH^- ions raises the conductivity. However, as the reaction progresses leading to more precipitate formation, the number of ions responsible for conductivity decreases. This critical balance between dissociated ions and precipitate formation dictates the overall conductivity of the solution.

Dissociation of Strong Electrolytes

The dissociation of strong electrolytes is a cornerstone of understanding conductivity in solutions. Strong electrolytes are substances that completely dissociate into ions when dissolved in water. Common examples include strong acids, such as hydrochloric acid (HCl), and strong bases like sodium hydroxide (NaOH), as well as many salts.

The complete dissociation is key for enabling the flow of electric current through the solution. Without free ions, there would be no charge carriers, and thus no electrical conductivity. In our exercise, magnesium sulfate (MgSO_4) and barium hydroxide (Ba(OH)_2) both dissociate completely, supplying a multitude of ions that initially boost conductivity. However, remember that the formation of insoluble precipitates during the reaction effectively removes ions from the solution, hence diminishing its conductive properties over time.