Question

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

Answer 1

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.

Answer 2

Answer : The pH of the solution is, 11.97

Solution :  Given,

Concentration (c) = 0.11 M

Base dissociation constant = $k_b=8.6\times 10^{-4}$

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 :

$k_b=\frac{[(CH_3CH_2)_2NH_2^+][OH^-]}{[(CH_3CH_2)_2NH]}$

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

$8.6\times 10^{-4}=\frac{(x)(x)}{(0.11-x)}$

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