Respuesta :
Answer : The concentration of [tex]H^+[/tex] ion, pH and pOH of solution is, [tex]1.05\times 10^{-5}M[/tex], 4.98 and 9.02 respectively.
Explanation : Given,
Concentration of [tex]OH^-[/tex] ion = [tex]9.5\times 10^{-10}M[/tex]
pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.
The expression used for pH is:
[tex]pH=-\log [H^+][/tex]
First we have to calculate the pH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (9.5\times 10^{-10})[/tex]
[tex]pOH=9.02[/tex]
The pOH of the solution is, 9.02
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-9.02=4.98[/tex]
The pH of the solution is, 4.98
Now we have to calculate the [tex]H^+[/tex] concentration.
[tex]pH=-\log [H^+][/tex]
[tex]4.98=-\log [H^+][/tex]
[tex][H^+]=1.05\times 10^{-5}M[/tex]
The [tex]H^+[/tex] concentration is, [tex]1.05\times 10^{-5}M[/tex]
Answer:
pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978
Explanation:
Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C
As, pOH = - log [OH⁻]
⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022
The self-ionisation constant of water is given by
Kw = [H⁺] [OH⁻] and pKw = pH + pOH
Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.
Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴
⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M
also,
pH + pOH = pKw = 14
⇒ pH = 14 - pOH = 14 - 9.022 = 4.978
