The pOH of an aqueous solution at 25°C was found to be 1.20. The pH of this solution is . The hydronium ion concentration is M. The hydroxide ion concentration is M.

pH = 12.80

[H3O+] = 1.58 * 10^-13 M

[OH-] = 0.063 M

Explanation:

Step 1: Data given

pOH = 1.20

Temperature = 25.0 °C

Step 2: Calulate pH

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.20 = 12.80

Step 3: Calculate hydronium ion concentration

pH = -log[H+] = -log[H3O+]

12.80 = -log[H3O+]

10^-12.80 = [H3O+] = 1.58 * 10^-13 M

Step 4: Calculate the hydroxide ion concentration

pOH = 1.20 = -log [OH-]

10^-1.20 = [OH-] = 0.063M

Step 5: Control [H3O+] and [OH-]

[H3O+]*[OH-] = 1* 10^-14

1.58 *10^-13 * 0.063 = 1* 10^-14

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante