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To 1.0 L of a 0.37 M solution of HClO2 is added 0.15 mol of NaF.

Calculate the [HClO2] at equilibrium.

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User Suvasis
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Okay, here are the steps to solve this problem:

1) Initially, the HClO2 solution has a concentration of 0.37 M. This means there are 0.37 moles of HClO2 per liter of solution.

2) We add 0.15 mol of NaF.

3) When NaF reacts with HClO2, they neutralize each other and form NaCl and FClO2. The balanced chemical equation is:

NaF + HClO2 → NaCl + FClO2

4) Since 1 mole of NaF reacts with 1 mole of HClO2, we can conclude that 0.15 mol of NaF will react with 0.15 mol of HClO2.

5) Therefore, after the reaction, there will be 0.37 - 0.15 = 0.22 moles of HClO2 left in the solution.

6) The volume of solution remains 1 L. So the new concentration of HClO2 at equilibrium is 0.22 M.

In summary:

Initial [HClO2]: 0.37 M

Moles of NaF added: 0.15 mol

Moles of HClO2 reacted: 0.15 mol

[HClO2] at equilibrium: 0.22 M

Does this make sense? Let me know if you have any other questions!

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User Shytikov
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