Question

You are given 100 mL of a solution of potassium hydroxide with a pH of 12.0. You are required to change the pH to 11.0 by adding water. How much water do you add?

Answer 1

NaOH ----> Na⁺ + OH^-

when pH is 12

pH = 12.0

pH = 14 - pOH

pH = 14 + log ([OH^- ] )

12 = 14 + log ([OH^- ] )

log ([OH^- ] ) = -2

[OH^- ] = exp(-2) = 0.135 mol/L

so 0.135 mole of NaOH in 1 L of water

when pH is 11

pH = 11.0

pH = 14 - pOH

pH = 14 + log ([OH^- ] )

11 = 14 + log ([OH^- ] )

log ([OH^- ] ) = -3

[OH^- ] = exp(-3) = 0.0.498 mol/L

0.0.498 mole in 1 L of water

amount of water to be present to make 0.135 mol/L of solution to 0.049 mol/L of solution = (0.135/0.0498) = 2.71

so we should (2.71 -1=) 1.71 L of water with the 1 L of water present with 0.135 mol solution as it will

make 0.135 mol / 2.71 L = 0.0498 mol/L

Answer

so we have to add 1.71 L of water to 12 pH solution of NaOH to make it 11 pH solution