Question

Part A

Calculate the molar concentration of OH− ions in a 7.4×10⁻² M solution of ethylamine (C2H5NH2)(Kb=6.4×10−4).

Express your answer using two significant figures.

[OH−] =		M

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Part B

Calculate the pH of this solution.

Express your answer using two decimal places.

pH =

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Answer 1

A)

C2H5NH2 dissociates as:

C2H5NH2 +H2O -----> C2H5NH3+ + OH-

7.4*10^-2 0 0

7.4*10^-2-x x x

Kb = [C2H5NH3+][OH-]/[C2H5NH2]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((6.4*10^-4)*7.4*10^-2) = 6.882*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

6.4*10^-4 = x^2/(7.4*10^-2-x)

4.736*10^-5 - 6.4*10^-4 *x = x^2

x^2 + 6.4*10^-4 *x-4.736*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.4*10^-4

c = -4.736*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.898*10^-4

roots are :

x = 6.569*10^-3 and x = -7.209*10^-3

since x can't be negative, the possible value of x is

x = 6.569*10^-3

so.[OH-] = x = 6.569*10^-3 M

Answer: 6.6*10^-3 M

2)

use:

pOH = -log [OH-]

= -log (6.569*10^-3)

= 2.1825

use:

PH = 14 - pOH

= 14 - 2.1825

= 11.8175

Answer: 11.82