Top topic this week
Back to FAQs
Related FAQs
Frequently Asked Question
Questions asked by users might not always be phrased in the clearest way.
/ a solution prepared from 0.055 mol of butanoic acid dissolved in sufficient rnwater to give 1.0 l of solution has a ph of 2.72. Determine ka for butanoic acid using an ICE table. The acid ionizes rnaccording to the following balanced equation.rnch3ch2ch2co2h (aq) + H2O (l) H3O+rn(aq) + CH3CH2CH2CO2rn-rn(aq
Accepted Answer
ICE Table:
| | CH3CH2CH2CO2H | H3O+ | CH3CH2CH2CO2-|
|:-----:|:-----------------:|:-------:|:-----------------:|
|Initial| 0.055 M | 0 | 0 |
|Change | -x | +x | +x |
|Equil| (0.055-x) M | x M | x M |
Ka Expression:
Ka = [H3O+][CH3CH2CH2CO2-] / [CH3CH2CH2CO2H]
Substitution and Calculation:
2.72 = 10^-pH = 10^-2.72 = 1.90 x 10^-3
Ka = (1.90 x 10^-3)^2 / (0.055 - 1.90 x 10^-3)
Ka = 1.74 x 10^-5
| | CH3CH2CH2CO2H | H3O+ | CH3CH2CH2CO2-|
|:-----:|:-----------------:|:-------:|:-----------------:|
|Initial| 0.055 M | 0 | 0 |
|Change | -x | +x | +x |
|Equil| (0.055-x) M | x M | x M |
Ka Expression:
Ka = [H3O+][CH3CH2CH2CO2-] / [CH3CH2CH2CO2H]
Substitution and Calculation:
2.72 = 10^-pH = 10^-2.72 = 1.90 x 10^-3
Ka = (1.90 x 10^-3)^2 / (0.055 - 1.90 x 10^-3)
Ka = 1.74 x 10^-5
Articles you might like