What is $pH$ if $(i)$ $[H_3O^+]=5.6xx10^-5molL^-1$ ; and $(ii)$ if $[HO^-]=10^-7molL^-1$ ?

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Answer 1 · 2017-06-25T20:06:50

$(i)$ $pH=4.25.................$

By definition, $pH=-log_10[H_3O^+]$ , and thus for (i)....

$pH=-log_10(5.6xx10^-5)=-(-4.25)=4.25$ (if I have counted up the zeroes properly........)

But since $[H_3O^+]xx[HO^-]=10^-14$ , we can take $log_10$ of both sides, and gets........

$log_10[H_3O^+]+log_10[HO^-]=log_(10)10^-14=-14$

Equivalently, multiplying each side by $-1$ ............

$underbrace(-log_10[H_3O^+])_color(red)(pH)underbrace(-log_10[HO^-])_color(blue)(pOH)=+14$

And thus our defining relationship......

$color(red)(pH)+color(blue)(pOH)= 14$

And so (ii) if $[HO^-]=10^-7*mol*L^-1$ , $pOH=7$ (why?), and $pH=7$ . Agreed?

What is pHpH if (i)(i) [H_3O^+]=5.6xx10^-5*mol*L^-1[H_3O^+]=5.6xx10^-5*mol*L^-1; and (ii)(ii) if [HO^-]=10^-7*mol*L^-1[HO^-]=10^-7*mol*L^-1?