Question

Sodium chloride is added to water (at 25°C) until it issaturated. Calculate the ${\mathbf{\text{Cl}}}^{\mathbf{-}}$concentration in such a solution.

Short answer

The$C{l}^{-}$ concentrationin the given solution is 6.15 M.

Step-by-step solution

Introduction to the Concept

The solubility product constant is a constant that determines the solubility equilibrium that exists between a substance in its solid-state and its ions in the aqueous state. It has the following general expression:

$A_x{B}_y\left(s\right)\iff x{A}^{+}\left(aq\right)+y{B}^{-}\left(aq\right)$

$K_{SP}={\left[A^{+}\right]}^x\times {\left[B^{-}\right]}^y$

where$K_{SP}$is the solubility product constant, and square brackets signify the concentration.

Determination of expression of S

The given NaCl dissolution in water is determined as:

$NaCl\left(s\right)\to {\text{Na}}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

The solubility product constant is expressed as:

$K_{SP}=\left[N{a}^{+}\right]\times \left[C{l}^{-}\right]$

The solubility s is expressed as:

$K_{SP}=\left(s\right)\times \left(s\right)$

$K_{SP}=s^2$

$s=\sqrt{K_{SP}}$

Determination of ΔGo

The following is the relationship between the solubility product constant $K_{SP}$and the standard free energy $\Delta G^o$.

$\Delta G^o=-RT\times ln\times K_{SP}-----\left(1\right)$

Where R = the universal gas constant

T= Temperature

Using the formula to calculate the standard free energy:

$\Delta G^o=\sum \Delta G^o\left(products\right)-\sum \Delta G^o\left(reactants\right)$

The standard free energy for the dissolving of NaCl is:

$\Delta G°=\left[\Delta G°\left(N{a}^{+}\left(aq\right)\right)+\Delta G°\left(C{l}^{-}\left(aq\right)\right)\right]-\left[\Delta G°NaCl\left(s\right)\right]$

Substituting the values is as follows:

$\Delta G°=\left[-262KJ/mol+\left(-131KJ/mol\right)\right]-\left[-384KJ/mol\right]$

$\Delta G°=-9KJ/mole$

$\Delta G°=-9\times {10}^{3}J/mol$

Determination of concentration of  Cl-

Temperature conversion from degrees Celsius to Kelvin:

T = 25 + 273

T = 298 K

Substituting T and $\Delta G^o$ values in equation (1),

$-9\times {10}^{3}J/mol=-(8.314J/molK)\times (298K)\times \ln \times K_{sp}$

$\ln \times K_{SP}=\frac{9\times {10}^{3}J/mol}{2477.572J/mol}$

$\ln \times K_{SP}=3.633$

$K_{SP}=e^{3.633}$

$K_{SP}=37.83$

Let’s substitute the above equation in solubility expression.

$s=\sqrt{K_{SP}}$

$s=\sqrt{37.83}$

$s=6.15$

Therefore the concentration of$C{l}^{-}$ in the given solution is 6.15 M.