Question

Complete the following table by calculating the missing entries and indicating whether the solution is acidic, basic or neutral. the pOH for one row is 9.75 and the H+ on the bottom row is 6.17x10^-11.

For the row that the pOH is 9.75 you have to find the pH, the H+, OH- and if its an acid or base. For the row that the H+ is 6.17 x 10^-11 you have to find the ph, pOH, OH- and if its an acid or base

Answer 1

Answer – We are given, First row, pOH = 9.75

We need to calculate the pH, H⁺ and OH^-. From the calculated H⁺ and OH^- we need to determine the solution is acidic or basic.

Calculation of pH

We know,

pH + pOH = 14

so, pH = 14 – pOH

            = 14 – 9.75

           = 4.25

Calculation of the H⁺

We know,

pH = -log [H⁺]

so, [H⁺] = 10^-pH_­

              = 10^-4.25

              = 5.62*10^-5 M

Calculation of [OH^-]

We know the ionic product

Kw = [H⁺][OH^-]

1.0*10^-14 = 5.62*10^-5 M * [OH^-]

[OH^-] = 1.0*10^-14 / 5.62*10^-5 M

          = 1.78*10^-10 M

From the above calculation we are getting [H⁺] > [OH^-] , so solution is acidic in nature.

Second row – we are given, [H⁺] = 6.17*10^-11 M.

we need to calculate the pH, pOH and OH^-. From the calculated H⁺ and OH^- we need to determine the solution is acidic or basic

Calculation of pH –

We know,

pH = -log [H⁺]

      = -log 6.17*10^-11 M

      = 10.2

Calculation of pOH

We know,

pH + pOH = 14

so, pOH = 14 – pH

            = 14 – 10.2

           = 3.79

Calculation of [OH^-]

We know the ionic product

Kw = [H⁺][OH^-]

1.0*10^-14 = 6.17*10^-11 M* [OH^-]

[OH^-] = 1.0*10^-14 / 6.17*10^-11 M

          = 1.62*10^-4 M

From the above calculation we are getting [H⁺] < [OH^-] , so solution is basic in nature.