Question

An aqueous solution contain 1.0x-⁹moles/dm³ of hydronium ions.Calculate the pOH of this solution.​

Answer 1

Answer:

Given [H

3

O

+

]=1×10

−10

M

At 25

∘

C,[H

3

O

+

][OH

−

]=10

−14

∴[OH

−

]=

10

−10

10

−14

=10

−4

Now, [OH

−

]=10

−p

OH

=10

−4

=10

−pOH

∴pOH=4

Answer 2

Answer:

The POH of this solution contains  $1\ *\ 10^{-9} \ \frac{moles}{dm^3}$ of hydronium ions is $6$·

Explanation:

Given that,

The concentration of hydronium ions,  $[H_3O^+]$  $=$  $1\ *\ 10^{-9} \ \frac{moles}{dm^3}$

To calculate the POH first, we need to find out the concentration of hydroxide ion·

For that, we know

The product of the concentration of hydronium ion and the hydroxide ion is equal to $K_W$·

That is,

   $[H_3O^+]\ [OH^-]\ =\ K_W$  $----(1)$

where,

$K_W$  is the water dissociation constant and its value is $1\ *\ 10^{-14} \ M^{2}$·

On modifying the equation No $1$ with the concentration of hydroxide ion term on the left-hand side, the equation becomes

   $[OH^-]\ =\ \frac{K_W}{[H_3O^+]}$  $----(2)$

On substituting the values, we get

⇒ $[OH^-]\ =\ \frac{1\ *\ 10^{-14} }{1\ *\ 10^{-9} }$

⇒ $[OH^-]\ =\ 10^{-6}$

Therefore,

The concentration of hydroxide ion in the solution is $10^{-6}$·

Now, to find out the POH we have

  POH $=$ $-$ log $[OH^-]$

On substituting the values, we get

⇒ POH $=$ $-$ log $[10^{-6} ]$

⇒ POH $=$ $6$

Hence,

The POH of this solution is $6$·