1. **Stating the problem:** Given concentrations, pH, and constants, we want to analyze the relationships between $C_1$, $C_2$, $n$, $K$, $pH_1$, $pH_2$, and conductivities $\sigma_1$, $\sigma_2$. 2. **Given data:** - $C_1 = 2.10^{-2}$ mol.L$^{-1}$ - $V = 200$ ml - $pH = 2.53$ - $K = 5.10^{-4}$ - $\sigma_1 = 0.124$ S.m$^{-1}$ 3. **Formulas:** - $C_2 = \frac{C_1}{n}$ - $n = \frac{K \cdot C_1 \cdot 10^{2pH_1}}{1 + K \cdot 10^{2pH_2}}$ - $pH_2 = pH_1 + \log \left( \frac{\sigma_1}{\sigma_2} \right)$ 4. **Step 1: Calculate $n$ using the formula:** $$n = \frac{K \cdot C_1 \cdot 10^{2pH_1}}{1 + K \cdot 10^{2pH_2}}$$ Since $pH_2$ depends on $pH_1$ and $\sigma_2$, we need to express $pH_2$ first. 5. **Step 2: Express $pH_2$:** $$pH_2 = pH_1 + \log \left( \frac{\sigma_1}{\sigma_2} \right)$$ 6. **Step 3: Substitute $pH_2$ into $n$:** $$n = \frac{K \cdot C_1 \cdot 10^{2pH_1}}{1 + K \cdot 10^{2 \left(pH_1 + \log \left( \frac{\sigma_1}{\sigma_2} \right) \right)}}$$ 7. **Step 4: Simplify the denominator:** Using the property $10^{a + b} = 10^a \cdot 10^b$: $$1 + K \cdot 10^{2pH_1} \cdot 10^{2 \log \left( \frac{\sigma_1}{\sigma_2} \right)}$$ 8. **Step 5: Simplify $10^{2 \log(x)}$:** Since $10^{\log(x)} = x$, then: $$10^{2 \log \left( \frac{\sigma_1}{\sigma_2} \right)} = \left( \frac{\sigma_1}{\sigma_2} \right)^2$$ 9. **Step 6: Final expression for $n$:** $$n = \frac{K \cdot C_1 \cdot 10^{2pH_1}}{1 + K \cdot 10^{2pH_1} \cdot \left( \frac{\sigma_1}{\sigma_2} \right)^2}$$ 10. **Step 7: Calculate $C_2$:** $$C_2 = \frac{C_1}{n}$$ **Summary:** - We expressed $n$ in terms of known quantities and $\sigma_2$. - To find numerical values, $\sigma_2$ or $pH_2$ must be known or measured. This completes the analysis of the given relations.