Answer:
Explanation:
Given;
pH of the solution = 4.33
H₃O+ is calculated as;
pH = -log[H₃O⁺]
4.33 = -log[H₃O⁺]
-4.33 = log[H₃O⁺]
10⁻⁴°³³ = [H₃O⁺]
[tex]Thus, [H_3O^+] = 1 *10^{-4.33}[/tex] M
[OH⁻] is calculated as;
[H₃O⁺] x [OH⁻] = 10⁻¹⁴
10⁻⁴°³³ x [OH⁻] = 10⁻¹⁴
[tex][OH^-] = \frac{10^{-14}}{10^{-4.33}} = 10^{-9.67}\\\\Thus, [OH^-] = 1*10^{-9.67} \ M[/tex]
pOH of the solution is calculated as;
pOH + pH = 14
pOH + 4.33 = 14
pOH = 14 - 4.33
pOH = 9.67
Thus, pOH of the solution is 9.67
Summary of the calculation:
