Question

In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 7.25 by using the green arrows adjacent to the pH value indicated on the probe in the solution. Once you adjust the pH, note the corresponding OH− ion concentration in M as given in the graphic on the left side of the simulation. Make sure to select the option "Concentration (mol/L)" above the graphic. Select on the Logarithmic scale below the graphic. Find the pOH of the solution.

Answer 1

Ans. The concentration of [H⁺] ions after setting pH to 7.25 is calculated as follow-

            pH = -log [H⁺]

            or, 7.25 = -log [H⁺]

            or, [H⁺] = antilog (-7.25)

            or, [H⁺] = 5.6 x 10^-8

Thus [H⁺] = 5.6 x 10^-8 M in the 0.5 L of neutral solution.

Now,

For any aqueous solution-

            [H⁺] [OH^-] = 10^-14

So, corresponding [OH^-] after simulation is given by-

            [OH^-] = 10^-14 / [H⁺] = 10^-14 / (5.6 x 10^-8) = 1.785 x 10^-7

Thus, corresponding [OH^-] after simulation = 1.785 x 10^-7

pOH = - log [OH^-] = -log (1.785 x 10^-7) = 6.75

Thus, pOH of simulated solution = 6.75

Note: (pOH = 14 – pH) can also be used to predict the pOH. However, its indicated to calculate pOH using [OH^-], so follow the steps.