Respuesta :
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
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
Answer:- [tex][H_3O^+]=4.2*10^-^1^1[/tex]
Solution:- Ammonia is a weak base. So, to calculate the hydroxide ion concentration we make the ice table:
[tex]NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)[/tex]
I 0.0032 0 0
C -X +X +X
E (0.0032-X) X X
[tex]K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Kb value for ammonia is [tex]1.8*10^-^5[/tex] . Let's plug in the values and solve for X.
[tex]1.8*10^-^5=\frac{X^2}{0.0032-X}[/tex]
Kb value is very low so we can neglect the X on the bottom.
[tex]1.8*10^-^5=\frac{X^2}{0.0032}[/tex]
On cross multiply:
[tex]X^2=1.8*10^-^5*0.0032[/tex]
[tex]X^2=5.76*10^-^8[/tex]
On taking square root:
[tex]X=2.4*10^-^4[/tex]
From ice table, [tex][OH^-]]=X[/tex]
So, [tex][OH^-]=2.4*10^-^4[/tex]
hydronium ion and hydroxide ion concentrations are related to each other by the formula:
[tex][H_3O^+][OH^-]=K_w[/tex]
where, Kw is the water dissociation constant and its value is [tex]1.0*10^-^1^4[/tex]
[tex][H_3O^+]=\frac{1.0*10^-^1^4}{2.4*10^-^4}[/tex]
[tex][H_3O^+]=4.2*10^-^1^1[/tex]
