Question

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH.


Part A) pH= 8.61

Part B) pH= 11.23

Part C) pH= 2.94


Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answer 1

Answer.

(a) (first part)

We all know at 25⁰C , pH + pOH = 14, pH =-log[H₃O⁺] and pOH =-log[OH^-]

pH = 8.61

so, pOH = (14 -8.61) =5.39

[H₃O⁺] = 10^(-8.61) = 2.45 x 10^-9 (M)

[OH^-] = 10^(-5.39) = 4.07 x 10^-6 (M)

(b) (second part)

We all know at 25⁰C , pH + pOH = 14, pH =-log[H₃O⁺] and pOH =-log[OH^-]

pH = 11.23

so, pOH = (14 -11.23) =2.77

[H₃O⁺] = 10^(-11.23) = 5.88 x 10^-12 (M)

[OH^-] = 10^(-2.77) = 1.69 x 10^-3 (M)

(c) (third part)

We all know at 25⁰C , pH + pOH = 14, pH =-log[H₃O⁺] and pOH =-log[OH^-]

pH = 2.94

so, pOH = (14 -2.94) =11.06

[H₃O⁺] = 10^(-2.94) = 1.14 x 10^-3 (M)

[OH^-] = 10^(-11.06) = 8.70 x 10^-12 (M)