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

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(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

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$< 7$

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(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 .......

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