To 1.0 L of a 0.34 M solution of HClO2 is added 0.15 mol of NaF....

To 1.0 L of a 0.34 M solution of HClO2 is added 0.15 mol of NaF. Calculate the [HClO2] at equilibrium.

PLEASE show ALL work!!

Expert Solution

no of moles of HClo2 = Molarity * volume in L

= 0.34*1 = 0.34 moles

HClO2 + NaF ----------> NaClO2 + H2O

I 0.34 0.15 0

C -0.15 -0.15 0.15

E 0.19 0 0.15

[HClO2] = 0.19moles

[HClO2] = 0.19/1 = 0.19 M >>>>> answer

queen_honey_blossom answered 1 year ago

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To 1.0 L of a 0.34 M solution of HClO2 is added 0.15 mol of NaF....

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