Question

He hydroxide ion concentration in household ammonia is 3.2 × 10−3 m at 25 °c. what is the concentration of hydronium ions in the solution

Answer 1

Answer:-
`[H_3O^+]=4.2*10^-^1^1`

Solution:- Ammonia is a weak base. So, to calculate the hydroxide ion concentration we make the ice table:

`NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)`

I 0.0032 0 0

C -X +X +X

E (0.0032-X) X X

`K_b=([NH_4^+][OH^-])/([NH_3])`

Kb value for ammonia is
`1.8*10^-^5` . Let's plug in the values and solve for X.

`1.8*10^-^5=(X^2)/(0.0032-X)`

Kb value is very low so we can neglect the X on the bottom.

`1.8*10^-^5=(X^2)/(0.0032)`

On cross multiply:

`X^2=1.8*10^-^5*0.0032`

`X^2=5.76*10^-^8`

On taking square root:

`X=2.4*10^-^4`

From ice table,
`[OH^-]]=X`

So,
`[OH^-]=2.4*10^-^4`

hydronium ion and hydroxide ion concentrations are related to each other by the formula:

`[H_3O^+][OH^-]=K_w`

where, Kw is the water dissociation constant and its value is
`1.0*10^-^1^4`

`[H_3O^+]=(1.0*10^-^1^4)/(2.4*10^-^4)`

`[H_3O^+]=4.2*10^-^1^1`

Answer 2

when Kw = [OH-][H3O+] ,

and when Kw is the ionic constant for water at 25°C = 1 x 10^-14

and we have [OH-] = 3.2 x 10^-3

this the concentration at equilibruim.

so by substitution:

1 x 10^-14 = 3.2 x 10^-3 * [H3O+] .

∴ [H3O] = (1 x 10^-14) / (3.2 x 10^-3)

= 3.125 x 10^-12 m.

so, the answer is: concentration of hydronium ions is = 3.125·10⁻¹² M