At 80,^{circ}C, distilled water has [H_3O^{+} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In distilled water,the auto-ionization reaction is $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq).$Since the water is neutral,the concentr

Vedclass · Accepted Answer

$1 	imes 10^{-12}$

Vedclass · Answer

(C) In distilled water,the auto-ionization reaction is $2H_2O(l) ightleftharpoons H_3O^{+}(aq) + OH^{-}(aq).$Since the water is neutral,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 1 	imes 10^{-6} \, mol/L.$The ionic product of water,$K_w,$ is defined as $K_w = [H_3O^{+}][OH^{-}].$Substituting the given values: $K_w = (1 	imes 10^{-6}) 	imes (1 	imes 10^{-6}) = 1 	imes 10^{-12}.$Therefore,the correct option is $C.$

At $80\,^{\circ}C,$ distilled water has $[H_3O^{+}]$ concentration equal to $1 \times 10^{-6} \, mol/L.$ The value of $K_w$ at this temperature will be

Similar Questions

What is the dissociation constant of water at $25^{\circ}C$?

The ionic product of water at $50^{\circ} C$ is $4 \times 10^{-14}$. The $pH$ and nature of water is :

Which equation represents the calculation of the ionic product for distilled water?

The $pH$ of a neutral solution in our body at normal temperature is ..........

What will be the sum of $pH$ and $pOH$ in an aqueous solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module