Question

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

Answer 1

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 $10^{-14}$

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 $10^{-14}$]

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

Answer 2

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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