At 90,^{circ}C,pure water has H_3O^{+} ion co | vedclass

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

Question

(C) In pure water,the dissociation is represented as $2H_2O \rightleftharpoons H_3O^{+} + OH^{-}$.Since the water is pure,the concentration of hydroni

Vedclass · Accepted Answer

$10^{-12}$

Vedclass · Answer

(C) In pure water,the dissociation is represented as $2H_2O ightleftharpoons H_3O^{+} + OH^{-}$.Since the water is pure,the concentration of hydronium ions is equal to the concentration of hydroxide ions: $[H_3O^{+}] = [OH^{-}] = 10^{-6}\,mol\,L^{-1}$.The ionic product of water $(K_w)$ is defined as $K_w = [H_3O^{+}][OH^{-}]$.Substituting the given values: $K_w = (10^{-6}) 	imes (10^{-6}) = 10^{-12}$.

At $90\,^{\circ}C$,pure water has $H_3O^{+}$ ion concentration of $10^{-6}\,mol\,L^{-1}$. The $K_w$ at $90\,^{\circ}C$ is

Similar Questions

If the $pH$ of water is less than ......,the water pipes undergo corrosion.

Which of the following will have the highest $pH$?

The $pH$ of a solution is $9.5$. The solution is

$A$ solution has $[H^{+}] = 0.001 \ M$. What is the value of $[OH^{-}]$?

If the value of the dissociation constant of water is $1.8 \times 10^{-16}$,what is the ionic product of water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module