Answer to Question #184299 in General Chemistry for dssd

Question

A weak base with a K_b value of 4 x 10^-8. The pH of a 0.01 M solution of the weak base is in the range of

• 7 to 8
• 4 to 5
• 9 to 10
• 11 to 12

(Explain why)

Answer 1

Range of 11 to 12.

The value of K_b

Kb

​ for caffine = $4\times 10^{-4}$

4×10−4

CafOH(aq)leftharpoons $Caf(aq)+OH^-(aq)$

CafOH(aq)leftharpoonsCaf(aq)+OH−

(aq)

Initial

  0     0.01 M   0

AT equilibrium:

 x     (0.01 -x)M   x

$K_b=\frac{x(0.01-x)}{(x)}$

$Kb ​=(x) x(0.01−x) ​ 4\times 10^{-4}=\frac{x(0.01-x)}{(x)} 4×10−4 =(x) x(0.01−x)$

​

Solving for x:

x = 0.0096 M

The pOH of the solution is given by :

$pOH=-\log[{OH^-}]$

pOH=-\log[x]

pOH=−log[x]

pOH=$-\log[0.0096]$

pOH=$-$ log[0.0096]

pOH = 2.02

pH= 14 $-$ pOH = 14 $-$ 2.02 = 11.98