If an aqueous solution of lead(II) nitrate (Pb(NO3)2) is mixed with an aqueous solution of potassium chloride (KCl) the precipit
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7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x ]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86