Question

52. pH of a neutral solution at 360 K is (Kw=3.6 10^-13 at 360 k)
(1) 7
(2) 7 + log 2
(3) 7 - log 6
(4) 7 + log 6
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Answer 1

The Ionic product of water is defined as below, It is the product of concentration of Hydronium ion and Hydroxy ion.

$K_{w} = [H _{3} O ^{ + } ] [OH ^{ - } ]$

For a neutral solution,

$[H _{3} O ^{ + } ] = [OH ^{ - } ]$

Let the concentration of Hydronium ion is x.

So, Ionic product of water

$K_{w} = [H _{3} O ^{ + } ] [OH ^{ - } ] \\ \\ K_{w} = {x}^{2}$

According to the question,

⇒x² = 3.6 × 10^ - 13

⇒x² = 36 × 10^ -14

⇒x = 6 × 10 ^ - 7

So, Concentration of Hydronium ion

$[H _{3} O ^{ + } ] = [OH ^{ - } ] = 6 \times {10}^{ - 7}$

Now, pH is the negative logarithm of the molar concentration of Hydronium ion.

$pH = - log(6 \times {10}^{ - 7} ) \\ \\ pH = - ( log(6) + log( {10}^{ - 7} ) ) \\ \\ pH = - ( - 7 + log(6) ) \\ \\ pH = 7 - log(6)$

Therefore, pH of a neutral solution at 360K is 7 - log6, Option 3