At 90,^oC,the pH of a 0.1,M NaCl aqueous solu | vedclass

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

Question

(A) At $25\,^oC$,the ionic product of water $(K_w)$ is $10^{-14}$,so $pH = -\log[H^+] = 7$ for a neutral solution. However,the dissociation of water i

Vedclass · Accepted Answer

$< 7$

Vedclass · Answer

(A) At $25\,^oC$,the ionic product of water $(K_w)$ is $10^{-14}$,so $pH = -\log[H^+] = 7$ for a neutral solution. However,the dissociation of water is an endothermic process. As the temperature increases to $90\,^oC$,the value of $K_w$ increases (it becomes greater than $10^{-14}$). Since $K_w = [H^+][OH^-]$,for a neutral solution $[H^+] = [OH^-] = \sqrt{K_w}$. At $90\,^oC$,$\sqrt{K_w} > 10^{-7}$,which means $[H^+] > 10^{-7}$. Therefore,$pH = -\log[H^+] Since $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,it does not undergo hydrolysis and the solution remains neutral. Thus,at $90\,^oC$,the $pH$ of a neutral $NaCl$ solution is less than $7$.

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

Similar Questions

If $[H_3O^{+}] = 3.5 \times 10^{-8} \ M$,calculate $[OH^{-}]$ and $pH$ of the solution.

Which of the following does not make any change in $pH$ when added to $10 \ mL$ of dilute $HCl$?

What is the $pH$ of a solution containing $2.2 \times 10^{-6} \ M$ hydrogen ions?

Hydrogen ion concentration in $mol/L$ in a solution of $pH = 5.4$ will be

The $pH$ of a sample of vinegar is $3.76$. Calculate the concentration of hydrogen ion in it.

Vedclass Test Series

Exam Paper Generator

Online Exam Module