Question

Using the following data, calculate the value of $K_{SP}$, for $Ba{\left(N{O}_3\right)}_2$, one of the least soluble of the common nitratesalts.

Short answer

The value of $K_{SP}$, for $Ba{\left(N{O}_3\right)}_2$, one of the least soluble of the common nitrate salts is $6.99\times {10}^{-4}$.

Step-by-step solution

Introduction to the Concept

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy, spontaneous change occurs.

The mathematical equation for the standard entropy value at room temperature is given as follows:

$\Delta G^o=\Delta {G^o}_{298}=\sum n{G^o}_{298}\left(Products\right)-\sum p{G^o}_{298}\left(Reactants\right)-----\left(1\right)$

In the balanced chemical equation, n and p indicate the coefficients of reactants and products.

$\Delta {G^o}_R=-RTlnK-----\left(2\right)$

Where

R = Universal gas constant ($R=8.314J/mol-k$)

T = Temperature

Determination of ΔGoR at 25oC and 1 atm

The dissolution reaction is:

$Ba{\left(N{O}_3\right)}_2\left(aq\right)\iff {\text{Ba}}^{2+}\left(aq\right)+2N{O}_3^{-}\left(aq\right)$

From equation (1),

$\Delta G_R^o=\left[1\Delta G_f^o\left(B{a}^{2+}\left(aq\right)\right)+2\Delta G_f^o\left(N{O}_3^{-}\left(aq\right)\right)\right]-\left[1\Delta G_f^o\left(Ba{\left(N{O}_3\right)}_2\left(aq\right)\right)\right]$

$$\begin{gathered} \Delta G_R^o=\left[1mole\times -561KJ/mol+2mole\times -109KJ/mol\right] \\ -\left[1mole\times -797KJ/mol\right] \end{gathered}$$

$\Delta G_R^o=-561-218+797$

$\Delta G_R^o=18KJ/mol$

Determination of K

From equation (2),

$\Delta G_R^o=-R\times T\times ln\times K_{SP}$

$K_{SP}=e^{-\frac{\Delta G_R°}{RT}}$

$=e^{-\frac{18KJ/mol}{0.08314KJ/mol-k\times 298k}}$

$K=6.99\times {10}^{-4}$

Therefore, the solubility of $Ba{\left(N{O}_3\right)}_2$is $6.99\times {10}^{-4}$.