posted by Fiona on .
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-]
note: pH +pOH = 14
Which becomes. 12.31 + pOH = 14,
POH = 14-12.31, = 1.69.
But pOH = -log[OH^-]