1. For a strong base, like NaOH and most of these, [OH-] is equal to the concentration of the base. So, if [NaOH] = 0.001 M, [OH-] = 0.001 M. pOH = - log[OH-], so for the first solution, pOH = 3. Since pH + pOH = 14, for that solution, pH = 11.
2. Ca(OH)2(aq) -------> Ca2+(aq) + 2OH-(aq)
1 mole of Ca(OH)2 produce 2 mole of OH-
0.01 M of Ca(OH)2 produce 0.02 M of OH-
pOH = -log[OH-]
pOH = -log(0.02) = 1.70
pH = 14 - pOH = 14 - 1.70 = 12.3
3. The pH of 10-8 N HCl can be calculated as follow:
[H+] = 10-8 N
As the solution is so dilute that water will also dissociate to produce H+ ions
H2O → H+ + OH-
Now, The Concentration of H+ released from water is 10-7 as the pH of water is equal to 7.
Hence the total concentration of H+ = 10-7 + 10-8
[H+] = 10-7 + 0.1 × 10-7
= 1.1 × 10-7
pH = -log [H+]
Hence pH = 6.95
Hence the pH of 10-8 N solution of HCl = 6.95.
4. pH = -log[H+],
pH = -log(2 x 10^-3) = 2.69897000434