Determine the [h3o+] of a 0.210 m solution of formic acid.

1 answer:

3 0

When the Pka for formic acid = 3.77
and Pka = -㏒ Ka
3.77 = -㏒ Ka
∴Ka = 1.7x10^-4

when Ka = [H+][HCOO-}/[HCOOH]

when we have Ka = 1.7x10^-4 &[HCOOH] = 0.21 m
so by substitution: by using ICE table value
1.7x10^-4 = X*X / (0.21-X)
(1.7x10^-4)*(0.21-X) = X^2 by solving this equation for X

∴X = 0.0059
∴[H+] = 0.0059
∴PH= -㏒ [H+]
= -㏒ 0.0059
= 2.23

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Arrange the following chloroarenes in increasing order of their reactivity in nucleophilic substitution to form their correspond

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The question is incomplete. Complete question is attached below
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Correct Answer: Option C i.e. I ~ III < IV < V < II

Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.

Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.

Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.

Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.

Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.