Question

If \(K_{s p}\) for \(\mathrm{HgSO}_{4}\) is \(6.4 \times 10^{-5}\), then solubility of this substance in mole per \(\mathrm{m}^{3}\) is : (a) \(8 \times 10^{-3}\) (b) \(6.4 \times 10^{-5}\) (c) \(8 \times 10^{-6}\) (d) None of these

Short answer

\(8 \times 10^{-3}\) mol/m^3

Step-by-step solution

Write down the dissociation equation

The first step is to write down the dissociation equation for \text{HgSO}_4. The dissociation of \text{HgSO}_4 in water can be represented as: \[ \text{HgSO}_4(s) \rightarrow \text{Hg}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]

Express the solubility product constant (Ksp)

Write the expression for the solubility product constant (Ksp) based on the dissociation equation: \[ K_{sp} = [\text{Hg}^{2+}][\text{SO}_4^{2-}] \] Since the stoichiometry of the dissociation is 1:1, we can assume that the solubility in mole per \text{m}^3 is 's' for both \text{Hg}^{2+} and \text{SO}_4^{2-}. Therefore, \[ K_{sp} = s \times s = s^2 \]

Calculate the solubility 's'

Now, solve for 's' using the given Ksp value: \[ 6.4 \times 10^{-5} = s^2 \] Taking the square root of both sides, we find: \[ s = \sqrt{6.4 \times 10^{-5}} \] \[ s = 8 \times 10^{-3} \, \text{mol/m}^3 \] This is the solubility of \text{HgSO}_4 in mol per \text{m}^3.

Key concepts

Dissociation Equation

The dissociation equation is fundamental in understanding how ionic compounds like HgSO₄ interact with solvents such as water. It represents the process where solid compounds break apart into their constituent ions when dissolved.

To illustrate, the dissolution of mercury sulfate can be represented as:
\[ \text{HgSO}_4(s) \rightarrow \text{Hg}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]
This equation indicates that one mole of solid HgSO₄ separates into one mole of mercury ions (Hg²⁺) and one mole of sulfate ions (SO₄^2-) when it dissolves. Recognizing this equimolar relationship is a crucial step in many calculations related to solubility and chemical equilibria.

Solubility Calculation

Solubility calculation involves determining the extent to which a substance can dissolve in a solvent at a given temperature and pressure. It's measured in terms of concentration, often moles per cubic meter (mol/m³). The solubility product constant (K_sp) provides a way to quantify this for sparingly soluble ionic compounds.

Using the given K_sp for HgSO₄, the solubility 's' in mol/m³ can be directly calculated:\[ K_{sp} = s^2 \]where 's' is the concentration of each ion in solution. Solving for 's' gives the solubility, which helps determine how much of the substance can be present in a saturated solution.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution to reach saturation. It's a key concept when dealing with substances that have limited solubility. For example, calculating the molar solubility of HgSO₄ involves finding the maximum amount of the compound that can dissolve in water without leaving any undissolved solid.

In this case, the calculated solubility, 8 x 10^-3 moles per cubic meter, represents the molar solubility of HgSO₄. This quantity informs us about the concentration at which the solution becomes saturated and no more HgSO₄ will dissolve.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a system are equal, resulting in no net change in the concentrations of reactants and products. It's a dynamic state where the ongoing dissolution and precipitation of a substance like HgSO₄ are balanced.

The solubility product constant, K_sp, plays a crucial role in understanding chemical equilibria for sparingly soluble compounds. It is the product of the molar concentrations of the ions at equilibrium and remains constant at a given temperature. When the product of the ion concentrations exceeds the K_sp, precipitation occurs; if it's lower, the solid will continue to dissolve until equilibrium is re-established.