Question

What is the pH of a solution where the [OH^- ] =1.52 x 10^-13 M?

What is the molarity of the OH^- ions in 0.040 M barium hydroxide, Ba(OH)₂?

Assume the barium hydroxide is fully dissociated into ions in the dilute solution.

What is the molarity of the OH^- ions in 0.040 M ammonia, NH₃?

What is the molarity of the OH^- ions in 0.040 M sodium hydroxide, NaOH?

What is the molarity of the H⁺ ions in 0.40 M nitric acid, HNO₃?

What is the molarity of the NO₃^- ions in 0.40 M nitric acid, HNO₃?

Answer 1

a) pOH = -log [OH^-]

           = -log (1.52*10^-13)

           = 12.82

    pH = 14-pOH

         = 14-12.82

         = 1.18

b) one mole of Ba(OH)2 contains two moles of OH- ions

concentration of OH- ions in 0.040 M Ba(OH)2 = 0.040 M *2

                                                                     = 0.080 M

c) one mole of NaOH contains one mole of OH- so

The molarity of the OH^- ions in 0.040 M sodium hydroxide, NaOH = 0.040 M

d) one mole of HNO3 contains one mole of H+ so

    The molarity of the H⁺ ions in 0.40 M nitric acid, HNO₃ = 0.40 M

e ) one mole of HNO3 contains one mole of NO3- so

     The molarity of the NO₃^- ions in 0.40 M nitric acid, HNO₃ = 0.40 M

To calculate the molarity of OH- ions in NH3 we need Kb value of NH3. please