The ionic product of water at 25,^{circ}C is | vedclass

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

Question

(B) The auto-ionization of water is an endothermic process represented by the equation: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.According to L

Vedclass · Accepted Answer

$1 	imes 10^{-12}$

Vedclass · Answer

(B) The auto-ionization of water is an endothermic process represented by the equation: $H_2O(l) ightleftharpoons H^+(aq) + OH^-(aq)$.According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature shifts the equilibrium to the right.Therefore,the concentration of $H^+$ and $OH^-$ ions increases,which leads to an increase in the ionic product of water $(K_w)$.Since $90\,^{\circ}C > 25\,^{\circ}C$,the value of $K_w$ at $90\,^{\circ}C$ must be greater than $10^{-14}$.Among the given options,only $1 	imes 10^{-12}$ is greater than $10^{-14}$.

The ionic product of water at $25\,^{\circ}C$ is $10^{-14}$. The ionic product at $90\,^{\circ}C$ will be

Similar Questions

What is the $pH$ of a solution having $H^{+}$ ion concentration $3.981 \times 10^{-7} \ M$? $(\log 3.981 = 0.6000)$

At $90\,^oC$,the $pH$ of a $0.1\,M$ $NaCl$ aqueous solution is .......

At $25\,^oC$,the $pH$ of a solution is $2$. If the $pH$ is doubled,the concentration of hydronium ions in the solution will ............

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

At $25 \, ^\circ C$,the $pH$ value of a solution is $6$. The solution is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module