SION 0
How many grams of NaCl are required to precipitate most of the Ag+ ions from 2.50x102
mL of 0.0113 M of AgNO3 solutions? Write the net ionic equation for the reaction.

The correct answer is 0.165 g NaCl.

Explanation:

The following is the precipitation reaction taking place between sodium chloride and silver nitrate:  

NaCl (aq) + AgNO₂ (aq) ⇒ NaNO₃ (aq) + AgCl (s)  

The complete ionic reaction of the reaction will be,  

Na⁺ + Cl⁻  + Ag⁺ + NO₃⁻ ⇒ AgCl (s) + Na⁺  + NO₃⁻

Hence, the net ionic equation for the mentioned reaction is:  

Ag⁺ (aq) + Cl⁻ (aq) ⇒ AgCl (s)

Thus, it can be witnessed that one mole of chloride ion is needed so that one mole of Ag⁺ ion get neutralized. There is a need to find the moles of silver ions present in the solution of AgNO₃ and then transform these moles to the moles of chloride ion. Ultimately, these moles can be converted to the concentration of sodium chloride needed.  

The no. of moles of silver ions found in silver nitrate solution is,  

(2.50 × 10² mL) × (0.0113 mol Ag⁺/1000 ml solution) = 2.83 × 10⁻³mol Ag⁺

Now the moles of chloride ions needed to precipitate the silver ions is,

(2.83 × 10⁻³ mol Ag⁺ ) × (1 mol Cl⁻/1 mol Ag⁺) = 2.825 × 10⁻³mol Cl⁻

The mass of sodium chloride needed for precipitating the silver ions will be,  

mass of NaCl = (2.83 × 10⁻³ mol Cl⁻) × (1 mol NaCl / 1 mol Cl⁻) × (58.44 grams / 1 mol NaCl)

= 0.165 gram NaCl.  

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