A sample of baking soda was dissolved in water and the pOH of the solution was found to be 5.81. Calculate the pH, [H₃O⁺] and [OH⁻] of the solution

Answer 1

Hello, Night,

At STP, pure water is in equilibrium with it's component ions:

H₂O = H⁺ + OH^- [Please excuse my laziness for using H⁺ instead of the politically correct H₃O⁺.

The equilibrium coefficient for this is 1x10^-14. So we can write

1x10^-14 =[H⁺]*[OH^-]

pOH is 5.81

[OH^-] = 10^-5.81

[OH^-] = 1.55x10^-6

Now we can find [H⁺]:

1x10^-14 =[H⁺]*[1.55x10^-6]

[H⁺] = [1x10^-14]/[1.55x10^-6]

[H⁺] = 6.45x10^-9M

The other approach would be to simply calculate pH from pOH:

pH + pOH = 14

pH = 14 - 5.81, or 8.19

[H⁺] = 10^-8.19

[H⁺] = 6.45x10^-9M



Bob