Question

calculate the ph of 0.005 ba(oh)2 aqueous solution

Answer 1

Ba (OH)2 is completely dissociated and so

Ba(OH)2---> Ba++ + 2OH-

so 0.005M yields 0.005*2=0.01 moles of OH-

So, pOH =log 1/[OH-] = log 1/10^-2= log 10^2 = 2

pH+pOH= 14

Answer= pH=14-2=12

Answer 2

Given:

Aqueous solution,

Ba(OH)2

Yields,

= 0.005 M

To find:

pH value = ?

Solution:

According to the question,

$Ba(OH)2--->Ba^{++}+2OH^-$

Moles of $OH^-$ will be:

= $0.005\times 2$

= $0.01$

We know that,

⇒ $pOH=log\frac{1}{OH^-}$

            $=log \frac{1}{10^2}$

            $=log 10^2$

            $=2$

hence,

The pH value will be:

⇒ $pH+pOH=14$

⇒             $pH=14-2$

⇒                   $=12$    

Thus the correct answer is "12".