Question

What is the ph of 0.11 m diethylamine, (ch3ch2)2nh, (kb = 8.6 × 10−4)?

Answer 1

Answer : The pH of the solution is, 11.97

Solution : Given,

Concentration (c) = 0.11 M

Base dissociation constant =
`k_b=8.6* 10^(-4)`

The equilibrium reaction for dissociation of
`(CH_3CH_2)_2NH` (weak base) is,

`(CH_3CH_2)_2NH+H_2O\rightleftharpoons (CH_3CH_2)_2NH_2^++OH^-`

initially conc. 0.11 0 0

At eqm. (0.11-x) x x

First we have to calculate the value of 'x'.

Formula used :

`k_b=([(CH_3CH_2)_2NH_2^+][OH^-])/([(CH_3CH_2)_2NH])`

Now put all the given values in this formula ,we get:

`8.6* 10^(-4)=((x)(x))/((0.11-x))`

By solving the terms, we get

`x=0.0093`

Thus, the concentration of hydroxide ion is:

`[OH^-]=x=0.0093M`

Now we have to calculate the pOH.

`pOH=-\log [OH^-]`

`pOH=-\log (0.0093)`

`pOH=2.03`

Now we have to calculate the pH.

`pH+pOH=14\\\\pH=14-pOH\\\\pH=14-2.03\\\\pH=11.97`

Therefore, the pH of the solution is, 11.97

Answer 2

Answer is: pH value of diethylamine is 11,96.
Chemical reaction: (CH₃CH₂)₂NH + H₂O → (CH₃CH₂)₂NH₂⁺ +OH⁻.
Kb((CH₃CH₂)₂NH) = 8,6·10⁻⁴.
c((CH₃CH₂)₂NH) = 0,11 M.
Kb((CH₃)₂NH) = c(OH⁻) · c((CH₃)₂NH₂⁺) ÷ c((CH₃)₂NH).
c(OH⁻) = c((CH₃CH₂)₂NH₂⁺) = x.
8,6·10⁻⁴ = x² ÷ (0,11 - x).
Solve quadratic equation: x = c(OH⁻) = 0,0092 M.
pOH = -log(0,0092 M) = 2,04.
pH = 14 - 2,04 = 11,96.