Question

If a solution of Ca(OH)2 has a pH of 12.31 how can I calculate the concentration of calcium hydroxide in mol L-1 ?

Write the dissociation equation:

Ca(OH)2 >>Ca+ + 2OH

So the concentration of the OH ion is twice the Ca(OH)2 .

Find the concentration of the OH ion


[H+]= antilog (-pH}

then [OH]= 1E-14/H+

Then [Ca(OH)2] = 1/2 [OH-]

Answer 1

note: pH +pOH = 14
Which becomes. 12.31 + pOH = 14,
POH = 14-12.31, = 1.69.
But pOH = -log[OH^-]
So 1.69=-log[OH^-]
1.69=log(1/OH)
Antilog(1.69)=1/OH
OH=1/antilog(1.69)