Question #4d832

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Answer 1 · 2017-03-16T22:40:38

$pOH=11.2$

We interrogate the equilibrium:

$2H_2O(l) rightleftharpoons H_3O^(+) + HO^(-)$

Now this equilibrium has been carefully measured under standard conditions, and at $298*K$ :

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

This is a mathematical relationship, we can divide, multiply or otherwise manipulate. One thing we can do is to take $-log_10$ OF BOTH SIDES,

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

But the RHS, $-log_(10)10^-14=-(-14)=14$ , by the very definition of the $"logarithmic function"$ . And thus,

$+14=-log_10[H_3O^+]-log_10[HO^-]$

And, $-log_10[H_3O^+]=pH$ , and $-log_10[HO^-]=pOH$ . This is simply by definition of the $p$ function, and thus,

$+14=pH+pOH$

And thus we are given that $[H_3O^+]=1.55xx10^-3*mol*L^-1$ , and so $pH=-(-2.81)=2.81.$

And so $pOH=11.19$

Capisce?