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
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