The aqueous solution having pH , 11 is how ma | vedclass

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(D) The $pH$ of a solution is defined as $pH = -\log[H^+]$.For a solution with $pH \, 8$,the concentration of hydrogen ions is $[H^+]_1 = 10^{-8} \, M

Vedclass · Accepted Answer

$1000$

Vedclass · Answer

(D) The $pH$ of a solution is defined as $pH = -\log[H^+]$.For a solution with $pH \, 8$,the concentration of hydrogen ions is $[H^+]_1 = 10^{-8} \, M$.For a solution with $pH \, 11$,the concentration of hydrogen ions is $[H^+]_2 = 10^{-11} \, M$.Since $pH + pOH = 14$ at $25^{\circ}C$,the $pOH$ values are:For $pH \, 8$,$pOH = 14 - 8 = 6$,so $[OH^-]_1 = 10^{-6} \, M$.For $pH \, 11$,$pOH = 14 - 11 = 3$,so $[OH^-]_2 = 10^{-3} \, M$.The ratio of basicity is the ratio of hydroxide ion concentrations: $\frac{[OH^-]_2}{[OH^-]_1} = \frac{10^{-3}}{10^{-6}} = 10^3 = 1000$.Therefore,the solution with $pH \, 11$ is $1000$ times more basic than the solution with $pH \, 8$.

The aqueous solution having $pH \, 11$ is how many times more basic than an aqueous solution having $pH \, 8$?

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