In a qualitative analysis, Bi³⁺$Bi^{3+}$ is detected by appearance of precipitate of BiO(OH)(s)$\mathrm{BiO(OH)(s)}$. Calculate pH when the following equilibrium exists at 298 K. BiO (OH) (s) leftharpoons BiO ^ + (aq) + OH ^ - (aq)$$\mathrm{BiO} (\mathrm{OH}) (\mathrm{s}) \rightleftharpoons \mathrm{BiO} ^ {+} (\mathrm{aq}) + \mathrm{OH} ^ {-} (\mathrm{aq})$$ K = 4 × 10⁻¹⁰$$K = 4 \times 10^{-10}$$ (Given: 2 = 0.3010$\log 2 = 0.3010$)

Core Logic

Titration: Oxalic acid (Weak Acid) against Sodium Hydroxide (Strong Base) in the burette. (Note: standard procedure often places base in the burette). Initially, phenolphthalein in the acidic solution (oxalic acid) is colourless. As base is added, pH increases. Near the equivalence point, the solution turns slightly alkaline. Phenolphthalein changes from colourless to pink in an alkaline medium (pH range 8.2 - 10.0).

Step 1: Final Conclusion

The colour change observed is from colourless to pink.

Pattern Recognition

Phenolphthalein: Acidic = Colourless. Basic = Pink. Weak Acid + Strong Base titration always has an alkaline equivalence point (pH > 7$pH > 7$), making phenolphthalein the ideal indicator.

Chapter Mix

Class 11 Chemistry: Equilibrium Class 11 Chemistry: Practical Chemistry

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