Question

calculate the Ph of 0.01 M ca(oh)2
Solution
log (2=0.3010)

Answer 1

What is the pH of 0.1m of CA(OH) 2?

Edward Willhoft, PhD, FIFST. from King's College London (1966)

Answered May 18, 2018 · Author has 2.7K answers and 1.9M answer views

You assume the ionic dissociation constant, K, of water.

K = [H+] x [OH-] = 10^-14 at 20 C.

Therefore,

[H+] = ( 10^-14)/[OH-]

O.1 m Ca(OH)2 gives 0.2 m of [OH-]

Therefore, substituting into above we have:

[H+] = (10^-14)/0.2

[H+] = 5 x 10^-14

Or in terms of pH:

-log [H+] = -log 5 + 14

pH = 14 - 0.698 = 13.302

The solublity of Ca(OH)2 is in fact considerably less than 0.1 mole per litre and more like 0.0233 mole at 20 C.

Therefore inserting the equivalent [OH-] concentration of 2 x 0.0233 moles into the above equation gives:

[H+] = (10^-14)/0.0466

From which the pH is 12.67 at the saturation concentration of Ca(OH)2 at 20 C.