What is the pH of 10-9 M HCl?
Here , [H+] for acid is < 10 -7 { which is not possible }
so,
total [H+] = 10 -7 + 10 -9
= 1.01 x 10 -7
Thus, pH = - log [H+]
= 6.99
The solubility of CaF2 in water at 18°C is 2.94 x 10-4 mole per liter. Calculate its solubility product.
CaF2 -----------> Ca++ + F2-
[ Ca++ ] = 2.94 x 10 -4
2 [ F- ] = 2 x 2.92 x 10 -4
thus , Ksp = [ Ca++] [ 2F-]2
= 4 x 2.94 x 10 -4 x ( 2.94 x 10 -4 ) 2
= 1.01 x 10 -10