Question: To 1.0 L of a .38 M solution of HClO2 is added to .20 mol of NaF. Calc

Question

Question: To 1.0 L of a .38 M solution of HClO2 is added to .20 mol of NaF. Calculate the HClO2 at equilibrium.

I know the answer is not going to .18 mol of HClO2, tried that one already. I believe it has something to do with the ice table.

  HClO2 + NaF ---> HF + NaClO2
I: .18 mol  0 mol     .2 mol .2 mol
C: x          x            x        x  
E: .18-x                 .2 -x .2-x

And I think Im stuck there, I keep getting .2 mol, which is the exact same thing I had... So that's useless. Any help would be appreciated/

Explanation / Answer

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.