what is the Ka for ethylammonium cation,C2H5NH3^+, at 25 celsius?(Kb for C2H5NH2=4.7x10^-4 at 25 Celsius)
2.12765957 x 10^-11
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Well, the Ka for the ethylammonium cation (C2H5NH3+) can be determined using the equation:
Ka = Kw / Kb
where Kw is the ionization constant of water (approximately 1.0 x 10^-14 at 25 degrees Celsius) and Kb is the base dissociation constant for ethylamine (C2H5NH2).
Now, since the Kw and Kb values are given, let's plug them into the equation:
Ka = (1.0 x 10^-14) / (4.7 x 10^-4)
After crunching the numbers, we get a Ka value of approximately 2.13 x 10^-11.
I hope that helps, but don't take my word for it. Ask a chemist, they always have a solution!
To find the Ka value for the ethylammonium cation, we can make use of the relationship between Ka and Kb for a conjugate acid-base pair.
The ethylammonium cation (C2H5NH3+) is the conjugate acid of the base ethylamine (C2H5NH2), for which the Kb value is given as 4.7x10^-4 at 25 degrees Celsius.
The relationship between Ka and Kb is expressed in the equation:
Ka x Kb = Kw
Where Kw is the ion product constant for water, which is equal to 1.0 x 10^-14 at 25 degrees Celsius.
Since we know the Kb value for ethylamine, we can calculate the corresponding Ka value for the ethylammonium cation as follows:
Ka x 4.7x10^-4 = 1.0x10^-14
Rearranging the equation:
Ka = (1.0x10^-14) / (4.7x10^-4)
Calculating the value:
Ka ≈ 2.13x10^-11
Therefore, the Ka value for the ethylammonium cation (C2H5NH3+) at 25 degrees Celsius is approximately 2.13x10^-11.