Answer
is: concentration of hydronium ions is 4.35·10⁻¹³ M.
[OH⁻]
= 2.3·10⁻² M; equilibrium
concentration of hydroxide ions.
The Kw (the ionic product for water) at
25°C is 1·10⁻¹⁴ mol²/dm⁶ or 1·10⁻¹⁴ M².
Kw = [H₃O⁺] · [OH⁻].
[H₃O⁺] = Kw ÷ [OH⁻].
[H₃O⁺] = 1·10⁻¹⁴ M² ÷ 2.3·10⁻² M.
[H₃O⁺] = 4.35·10⁻¹³ M.
Answer:
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
Explanation:
Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]
Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]
[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The ionic product of water is given as:
[tex]K_w=[H_3O^+][OH^-][/tex]
The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]
[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]
[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]
[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
