Question

Question #0f450

Answer 1

`K_b=1.8xx10^(-10)color(white)(x)"mol/l"`

Explanation:

We can firstly find the `pK_a` value using this expression for a weak acid:

`pH=1/2(pK_a-logc)`

Where `c` is the concentration of the acid.

`2.5=1/2pK_a-1/2log(0.18)`

`2.5=1/2pK_a+0.372`

`1/2pK_a=2.128`

`pK_a=4.256`

We now use the relationship:

`pK_a+pK_b=14`

`pK_b=14-4.256=9.744`

This means that `-logK_b=9.744`

From which `K_b=1.8xx10^(-10)color(white)(x)"mol/l"`