Question

. At 80°C, distilled water has (H.01 concentration
equal to 1 x 10-6 moleitre. The value of Kw at this
temperature will be
(1) 1 x 10-12
(2) 1* 10-15
(3) 1x 100
(4) 1 x 10​

Answer 1

Answer:

(1) 1×10-12

Explanation:

[H3O+] = [OH-] = 1 x 10-6 mole/litre

Kw = [H3O+] = [OH-] = [1 x 10-6] x [1 x 10-6]

= 1 x 10-12

Answer 2

The value of $K_{w}$ at given temperature is $1 \times 10^{-12}$.

Explanation:

It is known that for ionic product of water, the concentration of hydrogen ions is equal to the concentration of hydroxide ions.

That is,           $[H^{+}] = [OH^{-}]$

Since, it is given that the concentration of given hydrogen ions is $1 \times 10^{-6}$ mol liter.

And, the expression for relation between ionic product of water and concentration of hydrogen and hydroxide ions is as follows.

              $K_{w} = [H^{+}][OH^{-}]$

                      = $(1 \times 10^{-6}) \times (1 \times 10^{-6})$

                      = $1 \times 10^{-12}$

Thus, we can conclude that the value of $K_{w}$ at given temperature is $1 \times 10^{-12}$.

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