In the equilibrium equation,
HNO2 + H2O <--> H3O^+ + NO2^- , with 0.5M HNO2 and 0.5M NO2^-, which side of the reaction is favored? I am trying to set up an ice table, but can't figure out the change ( plus x or minus x) for each side. Please Help!
Chemistry Acid/Base Reaction Help?
2017-11-15 11:46 pm
回答 (2)
2017-11-16 1:24 am
Refer to: https://image.slidesharecdn.com/acidbasetable-151107033108-lva1-app6892/95/acid-base-table-1-638.jpg?cb=1446867098
Ka for HNO₂= 7.2 × 10⁻⁴
(You may find other values of Ka for HNO₂ from other sources.)
Consider the dissociation of HNO₂ :
____________ HNO₂(aq) __ ⇌ __ NO₂⁻(aq) __ + __ H⁺(aq) ___ Ka = 7.2 × 10⁻⁴
Initial: ________ 0.5 M _________ 0.5 M ________ 0 M
Change: _______ -y M __________ +y M _______ +y M
At eqm: _____ (0.5 - y) M _____ (0.5 + y) M ______ y M
(Since the initial [H⁺] is 0 M, then the change of [H⁺] must be positive because [H⁺] must not be negative.)
Due to the fact that Ka is small and common ion effect in the presence of 0.5 M NO₂⁻, the dissociation of HNO₂ would be to a very small extent.
It is assumed that 0.5 ≫ y, i.e.
[HNO₂] at equilibrium = (0.5 - y) M ≈ 0.5 M
[NO₂⁻] at equilibrium = (0.5 + y) M ≈ 0.5 M
Ka = [NO₂⁻] [H⁺] / [HNO₂]
7.2 × 10⁻⁴ = 0.5y / 0.5
y = 7.2 × 10⁻⁴
pH = -log[H⁺] = -log(7.2 × 10⁻⁴) = 3.14
Ka for HNO₂= 7.2 × 10⁻⁴
(You may find other values of Ka for HNO₂ from other sources.)
Consider the dissociation of HNO₂ :
____________ HNO₂(aq) __ ⇌ __ NO₂⁻(aq) __ + __ H⁺(aq) ___ Ka = 7.2 × 10⁻⁴
Initial: ________ 0.5 M _________ 0.5 M ________ 0 M
Change: _______ -y M __________ +y M _______ +y M
At eqm: _____ (0.5 - y) M _____ (0.5 + y) M ______ y M
(Since the initial [H⁺] is 0 M, then the change of [H⁺] must be positive because [H⁺] must not be negative.)
Due to the fact that Ka is small and common ion effect in the presence of 0.5 M NO₂⁻, the dissociation of HNO₂ would be to a very small extent.
It is assumed that 0.5 ≫ y, i.e.
[HNO₂] at equilibrium = (0.5 - y) M ≈ 0.5 M
[NO₂⁻] at equilibrium = (0.5 + y) M ≈ 0.5 M
Ka = [NO₂⁻] [H⁺] / [HNO₂]
7.2 × 10⁻⁴ = 0.5y / 0.5
y = 7.2 × 10⁻⁴
pH = -log[H⁺] = -log(7.2 × 10⁻⁴) = 3.14
2017-11-16 4:08 am
HNO2 (aq) + H2O (l) <======> H3O+ (aq) + NO2 - (aq)
** [ HNO2 ] = 0.5 M ; ** [ NO2-] = 0.5 M ; ** [ H3O+] = 0.5 M
The equilibrium expression : K = [ H3O+][ NO2-] / [ HNO2 ] = ( 0.5 x 0.5 ) / 0.5 = 0.5
** Since K < 1 : reactant - favored
** [ HNO2 ] = 0.5 M ; ** [ NO2-] = 0.5 M ; ** [ H3O+] = 0.5 M
The equilibrium expression : K = [ H3O+][ NO2-] / [ HNO2 ] = ( 0.5 x 0.5 ) / 0.5 = 0.5
** Since K < 1 : reactant - favored
收錄日期: 2021-04-18 17:57:05
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