Respuesta :
Answer is: the hydrogen
ion concentration is 2.5·10⁻⁷ M.
[OH⁻] = 4·10⁻⁸ mol/L, equilibrium concentration of hydroxide anion.
[H⁺] is the equilibrium concentration of hydrogen ions.
[OH⁻] · [H⁺] = 10⁻¹⁴ mol²/L², ionicproduct of water on room temperature.
[H⁺] = 10⁻¹⁴ mol²/L² ÷ 4·10⁻⁸ mol/L.
[H⁺] = 2,5·10⁻⁷ mol/L = 0,00000025 mol/L.
[OH⁻] = 4·10⁻⁸ mol/L, equilibrium concentration of hydroxide anion.
[H⁺] is the equilibrium concentration of hydrogen ions.
[OH⁻] · [H⁺] = 10⁻¹⁴ mol²/L², ionicproduct of water on room temperature.
[H⁺] = 10⁻¹⁴ mol²/L² ÷ 4·10⁻⁸ mol/L.
[H⁺] = 2,5·10⁻⁷ mol/L = 0,00000025 mol/L.
Answer: The pH of the solution is 6.68
Explanation:
To calculate the pOH of the solution, we use the equation:
[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex][OH^-]=4.8\times ^{-8}M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log (4.8\times 10^{-8})\\\\pOH=7.32[/tex]
Calculating the pH of the solution:
[tex]pH+pOH=14\\\\pH=14-7.32=6.68[/tex]
Hence, the pH of the solution is 6.68