State Ostwald's dilution law. What is the limitation of this law? Define the terms i) ionic product of water ii) common ion effect iii) degree of ionisation iv) P^H value. What will be the resultant…

Ostwald's Dilution Law

Statement: For a weak electrolyte that is partially ionized in solution, the degree of ionization ($\alpha$) is inversely proportional to the square root of its molar concentration ($C$).

$$\alpha \propto \frac{1}{\sqrt{C}}\text{or}\alpha =\sqrt{\frac{K}{C}}$$

Where $K$ is the ionization constant.

Limitation: This law is applicable only to weak electrolytes (e.g., acetic acid, ammonia) at low concentrations. It does not apply to strong electrolytes (e.g., $\mathrm{HCl}$, $\mathrm{NaCl}$) because they are almost completely ionized in dilute solutions, and their degree of ionization is effectively 1 regardless of dilution.

Acetic acidAcetic acid

Definitions

1. Ionic Product of Water ($K_w$): The product of the molar concentrations of hydrogen ions ($[\mathrm{H}{}^{+}]$) and hydroxide ions ($[\mathrm{OH}{}^{-}]$) in pure water or dilute aqueous solutions at a given temperature. At $25^{\circ }\text{C}$, $K_w=[\mathrm{H}{}^{+}][\mathrm{OH}{}^{-}]=1.0\times 10^{-14}$.

2. Common Ion Effect: The suppression of the ionization of a weak electrolyte by the addition of a strong electrolyte that shares a common ion. For example, adding $\mathrm{NaCl}$ to a solution of $\mathrm{CH}{}_3\mathrm{COOH}$ decreases the ionization of acetic acid due to the presence of common $\mathrm{Cl}{}^{-}$ ions (though typically demonstrated with $\mathrm{CH}{}_3\mathrm{COONa}$ adding $\mathrm{CH}{}_3\mathrm{COO}{}^{-}$).

3. Degree of Ionization ($\alpha$): The fraction of the total number of molecules of an electrolyte that dissociate into ions at a given time. $$\alpha =\frac{\text{Number of molecules ionized}}{\text{Total number of molecules present}}$$

4. ${\mathrm{P}}^{\mathrm{H}}$ Value: The negative base-10 logarithm of the molar concentration of hydrogen ions ($[\mathrm{H}{}^{+}]$) in a solution. $${\mathrm{P}}^{\mathrm{H}}=-{\log }_{10}[\mathrm{H}{}^{+}]$$

Calculation of Resultant ${\mathrm{P}}^{\mathrm{H}}$

Given:

• Volume of $\mathrm{HCl}$ solution, $V_1=200\text{ml}$
• ${\mathrm{P}}^{\mathrm{H}}$ of $\mathrm{HCl}$ solution = 2
• Volume of $\mathrm{NaOH}$ solution, $V_2=300\text{ml}$
• ${\mathrm{P}}^{\mathrm{H}}$ of $\mathrm{NaOH}$ solution = 12

Step 1: Calculate the concentration of $\mathrm{H}{}^{+}$ in the $\mathrm{HCl}$ solution. Since ${\mathrm{P}}^{\mathrm{H}}=-\log [\mathrm{H}{}^{+}]$, $$[\mathrm{H}{}^{+}{]}_{\mathrm{HCl}}=10^{-{\mathrm{P}}^{\mathrm{H}}}=10^{-2}\text{M}$$

Moles of $\mathrm{H}{}^{+}$ present: $$n_{\mathrm{H}{}^{+}}=[\mathrm{H}{}^{+}]\times V_1=10^{-2}\text{mol/L}\times 0.200\text{L}=2.0\times 10^{-3}\text{mol}$$

Step 2: Calculate the concentration of $\mathrm{OH}{}^{-}$ in the $\mathrm{NaOH}$ solution. First, find the ${\mathrm{P}}^{\mathrm{O}}\mathrm{H}$: $${\mathrm{P}}^{\mathrm{O}}\mathrm{H}=14-{\mathrm{P}}^{\mathrm{H}}=14-12=2$$ $$[\mathrm{OH}{}^{-}{]}_{\mathrm{NaOH}}=10^{-{\mathrm{P}}^{\mathrm{O}}\mathrm{H}}=10^{-2}\text{M}$$

Moles of $\mathrm{OH}{}^{-}$ present: $$n_{\mathrm{OH}{}^{-}}=[\mathrm{OH}{}^{-}]\times V_2=10^{-2}\text{mol/L}\times 0.300\text{L}=3.0\times 10^{-3}\text{mol}$$

Step 3: Determine the excess ion after neutralization. The reaction is: $$\mathrm{H}{}^{+}+\mathrm{OH}{}^{-}\to \mathrm{H}{}_2\mathrm{O}$$

Comparing moles:

• Moles of $\mathrm{H}{}^{+}=2.0\times 10^{-3}\text{mol}$
• Moles of $\mathrm{OH}{}^{-}=3.0\times 10^{-3}\text{mol}$

Since $n_{\mathrm{OH}{}^{-}}>n_{\mathrm{H}{}^{+}}$, the $\mathrm{H}{}^{+}$ is completely neutralized, and $\mathrm{OH}{}^{-}$ remains in excess.

Excess moles of $\mathrm{OH}{}^{-}$: $$n_{\mathrm{OH}{}^{-}}(\text{excess})=3.0\times 10^{-3}-2.0\times 10^{-3}=1.0\times 10^{-3}\text{mol}$$

Step 4: Calculate the concentration of excess $\mathrm{OH}{}^{-}$ in the final mixture. Total volume of the mixture, $V_{\text{total}}=V_1+V_2=200\text{ml}+300\text{ml}=500\text{ml}=0.500\text{L}$.

$$[\mathrm{OH}{}^{-}{]}_{\text{final}}=\frac{n_{\mathrm{OH}{}^{-}}(\text{excess})}{V_{\text{total}}}=\frac{1.0\times 10^{-3}\text{mol}}{0.500\text{L}}=2.0\times 10^{-3}\text{M}$$

Step 5: Calculate the ${\mathrm{P}}^{\mathrm{O}}\mathrm{H}$ and ${\mathrm{P}}^{\mathrm{H}}$ of the resultant solution. $${\mathrm{P}}^{\mathrm{O}}\mathrm{H}=-\log [\mathrm{OH}{}^{-}{]}_{\text{final}}=-\log (2.0\times 10^{-3})$$ $${\mathrm{P}}^{\mathrm{O}}\mathrm{H}=-(\log 2+\log 10^{-3})=-(0.3010-3)=3-0.3010=2.699$$

Finally, calculate ${\mathrm{P}}^{\mathrm{H}}$: $${\mathrm{P}}^{\mathrm{H}}=14-{\mathrm{P}}^{\mathrm{O}}\mathrm{H}=14-2.699=11.301$$

Result: The resultant ${\mathrm{P}}^{\mathrm{H}}$ of the mixture is 11.30.