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17 April, 12:54

If 91.9 mL of silver nitrate solution reacts with excess potassium chloride solution to yield 0.684 g of precipitate, what is the molarity of silver ion in the original solution?

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  1. 17 April, 14:39
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    0.052M

    Explanation:

    We'll begin by writing the balanced equation for the reaction. This is given below:

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

    Next, we shall determine the number of mole present in 0.684 g of precipitate, AgCl. This is illustrated below:

    Mass of AgCl = 0.684g

    Molar mass of AgCl = 108 + 35.5 = 143.5g/mol

    Number of mole of AgCl =.?

    Mole = Mass / Molar Mass

    Number of mole of AgCl = 0.684/143.5

    Number of mole of AgCl = 4.77*10¯³ mol

    Therefore, 4.77*10¯³ mol of AgCl was produced.

    Next, we shall determine the number of mole AgNO3 that reacted. This can be achieved by doing the following:

    From the balanced equation above,

    1 mole of AgNO3 reacted to produce 1 mole of AgCl.

    Therefore, 4.77*10¯³ mol of AgNO3 will also react to produce 4.77*10¯³ mol of AgCl.

    Next, we shall determine the molarity of AgNO3. This can obtain as follow:

    Volume = 91.9mL = 91.9/1000 = 0.0919L

    Mole of AgNO3 = 4.77*10¯³ mol

    Molarity =.?

    Molarity = mole / Volume

    Molarity = 4.77*10¯³ / 0.0919

    Molarity = 0.052M.

    Therefore, the molarity of the AgNO3 is 0.052M.

    Finally, we shall determine the molarity of silver ion in AgNO3. This can be obtained as follow:

    AgNO3 (aq) - > Ag + (aq) + NO3 - (aq)

    From the balanced equation above,

    1 mole of AgNO3 produced 1 mole of Ag+.

    Therefore, 0.052M AgNO3 will also produce 0.052M Ag+

    Therefore, the molarity of the silver ion, Ag + is 0.052M
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