Question

(i) What mass of Pb²⁺ ion is left in solution when 50 mL of 0.2 M Pb(NO₃)₂ is added to 50.0 mL of 1.5 M NaCl ? (K_spPbCl₂ = 1.7 × 10^–4) 

(ii) 0.16 g of N₂H₄ is dissolved in water and the total volume made up to 500 mL. Calculate the percentage of N₂H₄ that has reacted with water at this dilution. The K_b for N₂H₄ is 9.0 × 10^–6 M.

Answer 1

(i) Millimoles of Pb²⁺ before precipitation = 50 × 0.2 = 10 

Millimoles of Cl^– before precipitation = 50 × 1.5 = 75 

Assuming complete precipitation of PbCl₂ followed by establishment of equilibrium. 

Millimoles of Cl^– left after precipitation = 75 – 20 = 55 in 100 mL. 

After precipitation [Cl^–] = 0.55 M 

That means, we have to find out solubility of PbCl₂ in 0.55 M Cl^– ion solution.