At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00 B) 7.27 C) 6.14 D) 6.63 E) 7.86

The pOH of the solution is 7.85.

We know that the ion product of water is given as;

[H^+] × [OH^-] =  51.3*10^-14

This value kw is given as  51.3*10^-14.

Hence;

Given that [H^+] = [OH^-] =  [H^+]^2

Then;

[H^+]^2 = 51.3*10^-14

[H^+] = √51.3*10^-14

[H^+] = 7.16 × 10^-7

pH = - log[7.16 × 10^-7]

pH = 6.15

Now;

pH + pOH = 14

pOH = 14 - 6.15

pOH = 7.85

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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

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