Respuesta :
q1)
pH scale is used to determine how acidic a solution is
pH can be calculated as follows
pH = - log[H₃O⁺]
when the pH has been given , the hydronium ion concentration can be determined
[H₃O⁺] = antilog(-pH)
[H₃O⁺] = antilog(-4.65)
[H₃O⁺] = 2.24 x 10⁻⁵ M
[H₃O⁺] is 2.24 x 10⁻⁵ M
q2)
pOH scale is used to determine the alkalinity of a solution. pOH can be calculated using the OH⁻ ion concentration
pOH = -log[OH⁻]
OH⁻ ion concentration can be calculated knowing the pOH value
[OH⁻] = antilog(-pOH)
[OH⁻] = antilog(-4.65)
[OH⁻] = 2.24 x 10⁻⁵ M
the answer is 2.24 x 10⁻⁵ M
pH scale is used to determine how acidic a solution is
pH can be calculated as follows
pH = - log[H₃O⁺]
when the pH has been given , the hydronium ion concentration can be determined
[H₃O⁺] = antilog(-pH)
[H₃O⁺] = antilog(-4.65)
[H₃O⁺] = 2.24 x 10⁻⁵ M
[H₃O⁺] is 2.24 x 10⁻⁵ M
q2)
pOH scale is used to determine the alkalinity of a solution. pOH can be calculated using the OH⁻ ion concentration
pOH = -log[OH⁻]
OH⁻ ion concentration can be calculated knowing the pOH value
[OH⁻] = antilog(-pOH)
[OH⁻] = antilog(-4.65)
[OH⁻] = 2.24 x 10⁻⁵ M
the answer is 2.24 x 10⁻⁵ M
These are two questions and two answers:
1) What is the hydronium ion concentration in a solution of HCl that has a pH of 4.65?
i) Data:
pH = 4.65
[H₃O⁺] = ?
ii) Formula: pH = - log [H₃O⁺]
iii) Calculations:
4.65 = - log [H₃O⁺] ⇒ [H₃O⁺] =
[tex][H_3O^+]=10^{-4.65}=2.24.10^{-5}[/tex]
2) What is the hydroxide ion concentration in a solution of NH3 with a pOH of 4.65?
i) Data:
pOH = 4.65
[OH⁻] = ?
ii) Formula: pOH = - log [OH⁻]
iii) Calculations:
4.65 = - log [OH⁻] ⇒ [OH] =
[tex][OH^-]=10^{-4.65}=2.24.10^{-5}[/tex]
1) What is the hydronium ion concentration in a solution of HCl that has a pH of 4.65?
i) Data:
pH = 4.65
[H₃O⁺] = ?
ii) Formula: pH = - log [H₃O⁺]
iii) Calculations:
4.65 = - log [H₃O⁺] ⇒ [H₃O⁺] =
[tex][H_3O^+]=10^{-4.65}=2.24.10^{-5}[/tex]
2) What is the hydroxide ion concentration in a solution of NH3 with a pOH of 4.65?
i) Data:
pOH = 4.65
[OH⁻] = ?
ii) Formula: pOH = - log [OH⁻]
iii) Calculations:
4.65 = - log [OH⁻] ⇒ [OH] =
[tex][OH^-]=10^{-4.65}=2.24.10^{-5}[/tex]