Question

What is the ph of a solution obtained by dissolving 0.0005 mole of strong electrolyte, ca(oh)2 to form 100 ml of a saturated aqueous solution?

Answer 1

pH of the solution is 12

Explanation:

concentration of Ca(OH)2= 0.0005 * 100 /100

= .0005 mol/l

Ca(OH)2 ............> Ca^+2 + 2 OH^-

initially. 1. 2

.0005 .0005 * 2

= .o1

pOH= -log( OH) = -log(.01) = 2

pH = 14 - 2

= 12

I hope this is helpful to you...

Answer 2

Answer:12

Explanation:

Given:n=0.0005 mol

V=100ml/10^-1 lit

Ph =?

(OH-) = normality

=molarity ×acidity

=5×10^-3×2

=10^-2

(H+) = Kw/(oh-)

=10^-12

Ph= - log (H+)

= - log (10^-12)

=Ph =12