Respuesta :
Explanation:
Let us assume that the concentration of [[tex]OH^{-}[/tex] and [tex]H^{+}[/tex] is equal to x. Then expression for [tex]K_{w}[/tex] for the given reaction is as follows.
[tex]K_{w} = [OH^{-}][H^{+}][/tex]
[tex]K_{w} = x^{2}[/tex]
[tex]6.8 \times 10^{-15} = x^{2}[/tex]
Now, we will take square root on both the sides as follows.
[tex]\sqrt{6.8 \times 10^{-15}} = \sqrt{x^{2}}[/tex]
[tex][H^{+}] = 8.2 \times 10^{-8}[/tex] M
Thus, we can conclude that the [tex]H_{3}O^{+}[/tex] concentration in neutral water at this temperature is [tex]8.2 \times 10^{-8}[/tex] M.
Answer: The concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
Explanation:
The chemical equation for the ionization of water follows:
[tex]2H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The expression of [tex]K_w[/tex] for above equation, we get:
[tex]K_w=[H_3O^+]\times [OH^-][/tex]
We are given:
[tex]K_w=6.8\times 10^{-15}[/tex]
[tex][H^+]=[OH^-]=x[/tex]
Putting values in above equation, we get:
[tex]6.8\times 10^{-15}=x\times x\\\\x=8.2\times 10^{-8}M[/tex]
Hence, the concentration of [tex]H_3O^+[/tex] in neutral water is [tex]8.2\times 10^{-8}M[/tex]
