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Recalculate the percent dissociation of 0.19M HN3 in the presence of 0.19M HCl.

Recalculate the percent dissociation of 0.19M HN3 in the presence of 0.19M HCl.

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Expert Solution

Recalculate the percent dissociation of 0.19M HN3 in the presence of 0.19M HCl.

   Given : Molarity of HN3 = 0.19 M , [HCl]= 0.19 M

We know HCl is strong acid so it ionizes completely.

The[ H3O+ ]= [HCl]= 0.19 M

To show the dissociation of HN3, we have to show its reaction with water. and then by using ICE chart we have to find final concentration of H3O+ (at equilibrium )

                 HN3 (aq) +   H2O (l)    -----> H3O+ (aq)     +    N3- (aq)

I            0.19                                              0.19                    0

C          -x                                                  +x                         +x

E          (0.19-x)                                      (0.19 +x)                  x

We look for ka value for HN3 in reference table.

Ka = 1.9E-5

Equilibrium constant expression.

Ka = (1.9E-5) = (x+0.19) x / (0.19-x)

Now we have t solve for x

x in the denominator and numerator is neglected by considering 5% approximation

1.9 E-5 = 0.19 x / 0.19

x = 1.9 E-5

Percent dissociation = x/ [HN3] * 100

= 1.9E-5 / 1.9 * 100

= 1 E-5 * 100 = 1E-3 %

Or 0.001 %

queen_honey_blossom answered 6 hours ago
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