Answer:
0.00000223
Explanation:
pKa for C2H5NH3+ = 10.7
pKw = 14.0
pKa + pKb = pKw
10.7 + pKb = 14.0
pKb = 14.0 - 10.7
pKb = 3.30
C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce C2H5NH3^+ + Cl^-
The base hydrolysis reaction: C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)
This reaction is described by Kb.
Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]
Let [C2H5NH2] = [H3O^+] = x,
so [C2H5NH3^+] = 0.26 - x
Kb = x^2/(0.26 - x) = 2.00 x 10^-11
Let's solve for x. In this equation, It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x is approximately equal to 0.26. We won't know until we do the calculation.
We get: x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0
With the use of a quadratic calculator.
x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]
0.26 - x is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.
[C2H5NH3^+] = 0.26 M = [Cl^-]
NOTE:
pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65
Ka is the acid dissociation constant
Kb is the base dissociation constant
The concentration of each specie can be obtained using the ICE table.
The term Ka is called the acid dissociation constant. It shows the extent to which an acid is dissociated in solution.
We have to set up the ICE table as follows;
C2H5NH3^+ + H2O ⇄ C2H5NH2 + H3O^+
I 0.26 0 0
C -x +x +x
E 0.26 - x x x
But Ka = [C2H5NH2] [H3O^+]/[C2H5NH3^+]
But Ka = 2.3×10–11
2.3×10–11 = x^2/0.26 - x
2.3×10–11 (0.26 - x) = x^2
5.98 * 10^-12 - 2.3 * 10^-11x = x^2
x^2 + 2.3 * 10^-11x - 5.98 * 10^-12 = 0
x= 0.00000245 M
Hence;
[C2H5NH2] = 0.00000245 M
[H3O^+] = 0.00000245 M
[C2H5NH3^+] = 0.26 - 0.00000245 = 0.259 M
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