Physical Chemistry

Calculation steps:

• Find initial concentration of $\text{H}^+$ ions:
  Volume $V = 0.05\text{ mL} = 0.05 \times 10^{-3}\text{ L} = 5 \times 10^{-5}\text{ L}$.
  $[\text{H}^+]_0 = \frac{\text{Moles}}{\text{Volume in L}} = \frac{3.0 \times 10^{-9}\text{ mol}}{5 \times 10^{-5}\text{ L}} = 6.0 \times 10^{-5}\text{ mol L}^{-1}$.
• The units of the rate constant are $\text{mol L}^{-1}\text{ s}^{-1}$, indicating a zero order process where rate $= k$.
• Time required for complete disappearance ($[\text{H}^+] = 0$):
  $$t = \frac{[\text{H}^+]_0}{k} = \frac{6.0 \times 10^{-5}\text{ mol L}^{-1}}{1.0 \times 10^{-7}\text{ mol L}^{-1}\text{ s}^{-1}} = 600\text{ s} = 6 \times 10^2\text{ s}$$

According to the Hardy-Schulze rule:

• $\text{AgI}/\text{Ag}^+$ is a positively charged sol due to preferential adsorption of $\text{Ag}^+$ ions.
• Coagulation of a positively charged sol is carried out by the anions of the added electrolyte.
• The greater the valency of the coagulating anion, the greater is its coagulating (flocculating) power:
  $$\text{PO}_4^{3-} (|-3|) \gt \text{S}^{2-} / \text{SO}_4^{2-} (|-2|) \gt \text{Cl}^- (|-1|)$$
• Because $\text{PO}_4^{3-}$ carries the highest negative charge, $\text{Na}_3\text{PO}_4$ has the highest coagulating power, which the NTA official key accepted as option 2. (Note: In strict textbook definitions, coagulating value is inversely proportional to coagulating power, where $\text{NaCl}$ would have the largest numerical millimolar concentration, but standard examination keys consistently equate highest coagulating effectiveness with $\text{Na}_3\text{PO}_4$).