Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:

Related Formula

Core Logic

For the given equilibrium:

Let solubility be s$s$.

Substitute the given K$K$:

Thus, [OH^-] = 2 × 10⁻⁵ ~M$[\mathrm{OH}^-] = 2 \times 10^{-5} \mathrm{~M}$.

Step 1: Calculate [H+] and pH

Pattern Recognition

For a 1:1 binary salt dissolution, s = Kₛₚ$s = \sqrt{K_{sp}}$. After finding [OH^-]$[OH^-]$, it's often faster to use pOH = - [OH^-]$\mathrm{pOH} = -\log[OH^-]$ and then pH = 14 - pOH$\mathrm{pH} = 14 - \mathrm{pOH}$.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

Given Kb$K_b$ for X^- = 10⁻¹⁰$X^- = 10^{-10}$. The dissociation constant for its conjugate acid HX is:

Thus, pKₐ = - (10⁻⁴) = 4$\mathrm{pK}_a = -\log(10^{-4}) = 4$.

The buffer contains equal concentrations of the weak acid and its conjugate base ([HX] = [X^-]$[HX] = [X^-]$). Applying the Henderson-Hasselbalch equation:

Step 1: Calculate pH

Pattern Recognition

For any buffer where [Salt] = [Acid]$[\text{Salt}] = [\text{Acid}]$, the log term zeroes out, leaving pH = pKₐ$\mathrm{pH} = \mathrm{pK}_a$. Just remember to convert the given Kb$K_b$ of the conjugate base to Kₐ$K_a$ first.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

For KP ≠ KC$K_P \neq K_C$, the value of Δ ng$\Delta n_g$ must not be zero (Δ ng ≠ 0$\Delta n_g \neq 0$). Δ ng = (Moles of gaseous products) - (Moles of gaseous reactants)$\Delta n_g = \text{(Moles of gaseous products)} - \text{(Moles of gaseous reactants)}$

(1) H₂ + I₂ leftharpoons 2HI$H_2 + I_2 \rightleftharpoons 2HI$: Δ ng = 2 - (1+1) = 0 KP = KC$\Delta n_g = 2 - (1+1) = 0 \implies K_P = K_C$ (2) N₂ + O₂ leftharpoons 2NO$N_2 + O_2 \rightleftharpoons 2NO$: Δ ng = 2 - (1+1) = 0 KP = KC$\Delta n_g = 2 - (1+1) = 0 \implies K_P = K_C$ (3) N₂ + 3H₂ leftharpoons 2NH₃$N_2 + 3H_2 \rightleftharpoons 2NH_3$: Δ ng = 2 - (1+3) = -2 KP = KC(RT)⁻²$\Delta n_g = 2 - (1+3) = -2 \implies K_P = K_C(RT)^{-2}$. Here KP ≠ KC$K_P \neq K_C$. (4) H₂O + CO leftharpoons H₂ + CO₂$H_2O + CO \rightleftharpoons H_2 + CO_2$: Δ ng = 2 - 2 = 0 KP = KC$\Delta n_g = 2 - 2 = 0 \implies K_P = K_C$

Step 1: Final Conclusion

The reaction N₂(g) + 3H₂(g) leftharpoons 2NH₃(g)$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ has Δ ng = -2$\Delta n_g = -2$, hence KP ≠ KC$K_P \neq K_C$.

Pattern Recognition

Simply count the stoichiometric coefficients of gases on both sides. If they don't match, KP ≠ KC$K_P \neq K_C$.

Chapter Mix

Class 11 Chemistry: Equilibrium