What is the pH of a 0.000460 M solution of $C a {(O H)}_{2}$ $Ca(OH)_2$ ?

Question

23722 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-21T07:11:30

Approx. $11$ $11$

$p O H$ $pOH$ $=$ $=$ $- {log}_{10} [H O^{-}]$ $-log_10[HO^-]$ .

Thus $p O H = - {log}_{10} {0.009200}$ $pOH=-log_10{0.009200}$ $=$ $=$ $- (- 3.04) = 3.04$ $-(-3.04)=3.04$

But it is a FACT that $p H + p O H = 14$ $pH +pOH=14$

Thus $p H ≅ 11$ $pH~=11$

Why is the concentration of $H O^{-}$ $HO^-$ double that of $C a {(O H)}_{2}$ $Ca(OH)_2$ ?

PS If you have trouble with the $log$ $log$ function, voice your problem . Someone here will help you.

What is the pH of a 0.000460 M solution of Ca(OH)_2Ca(OH)2Ca(OH)_2?