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22 April, 03:46

If 67.1 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.235 g of precipitate, what is the molarity of silver ion in the original solution?

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  1. 22 April, 05:43
    0
    0.098 M

    Explanation:

    The equation for the reactions is given by;

    AgNO₃ (aq) + KCl (aq) → AgCl (s) + KNO₃ (aq)

    Volume of silver nitrate is 67.1 mL Mass of the precipitate, AgCl is 0.235 g

    We are required to calculate the molarity of silver ion in the original solution;

    Step 1: Calculate the moles of AgCl

    We know that, Moles = Mass : Molar mass

    Molar mass of AgCl = 143.32 g/mol

    Therefore;

    Moles of AgCl = 0.235 g : 143.32 g/mol

    = 0.00164 moles

    Step 2: Moles of silver nitrate

    From the equation, 1 mole of silver nitrate reacts to yield 1 mole of AgCl

    Therefore, Moles of silver nitrate = moles of AgCl

    Thus, moles of silver nitrate = 0.00164 moles

    Step 3: Moles of silver ions

    Moles of silver nitrate = 0.00164 moles

    But;

    AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq)

    Therefore, moles of Ag⁺ is equal to the moles of AgNO₃

    Moles of Ag⁺ = 0.00164 moles

    Step 4: Molarity of silver ions in silver nitrate

    Molarity = Moles : Volume

    Therefore;

    Molarity of Ag⁺ = 0.00164 moles : 0.0671 L

    = 0.0982 Moles/L

    = 0.098 M

    Therefore, the molarity of silver ions in silver nitrate is 0.098 M
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