1. what would you predict for the rate if 2.5 ml of water were mixed with...

1. what would you predict for the rate if 2.5 ml of water were mixed with 2.5 ml of potassium iodate solution, then 2 ml of the sodium meta bisulfite. please explain? 2. pleas calculate the molerity. when 10 ml of 0.05 M sodium meta-bisulfite solution mixed with 10 ml of the starch solution.

Expert Solution

In this reaction, potassium iodate and sodium metabisulfite react to form iodine. The starch solution serves as an indicator of the end of the reaction by forming a deep-blue colored starch–iodine complex. The reaction time can thus be measured by noting the time until the appearance of the blue color for each trial. The overall reaction occurs in a series of steps, as outlined below.

Step 1: Formation of iodinate ions (IO₃^-) and hydrogen sulfite ions (HSO₃^-)

Solution A: KIO_3(aq) à IO₃^-_(aq) + K⁺_(aq)

Solution B: H₂O_(l) + Na₂S₂O_5(s) à 2HSO₃^-_(aq) + 2Na⁺_(aq)

Step 2: Iodate ions react with hydrogen sulfite ions to produce iodide ions (I^-)

IO₃^-_(aq) + 3HSO₃^-_(aq) à I^-_(aq)+ 3H⁺_(aq)+ 3SO₄^2-_(aq)

Step 3: In the presence of hydrogen ions (H⁺), the iodide ions react with excess iodate ions to produce iodine (I₂). This is the kinetically slow step in the overall reaction.

6H⁺_(aq) + 5I^-_(aq) + IO₃^-_(aq) à 3I_2(aq) + 3H₂O_(l)

Step 4: Before the iodine can react with the starch to produce a dark blue colored complex, it immediately reacts with any hydrogen sulfite ions still present to form iodide ions.

I_2(aq) + HSO₃^-_(aq) + H₂O_(l) à 2I^-_(aq)+ 3H⁺_(aq)+ SO₄^2-_(aq)

Step 5: Once all of the hydrogen sulfite ions have reacted, the iodine is then free to react with the starch to form the familiar dark-blue colored complex.

I_2(aq) + starch dark-blue colored complex

queen_honey_blossom answered 2 years ago

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