Represent the autoprotolysis of water, and explain how $pH$ defines the acidity of the solution. How are $pH$ and $pOH$ defined?

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Answer 1 · 2017-09-04T15:35:45

$5.$ $2H_2OrightleftharpoonsH_3O^+ + HO^-$

Note that $[H_3O^+]$ is synonymous with $[H^+]$

$2H_2OrightleftharpoonsH_3O^+ + HO^-$

And $K_w=[H_3O^+][HO^-]=10^-14$ at $298*K$ . This equilibrium has been extensively measured.

$"6. and 7"$ We can take $log_10$ of both sides...........

$log_10K_w=log_10[H_3O^+]+log_10[HO^-]$

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

And so on rearrangement.........

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

But by definition........

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

And so $14=pH+pOH$ , under standard conditions of $298*K$ and near $1*atm$ .

If $pH=7$ , then, clearly, $[H_3O^+]=[HO^-]$ , and the solution is NEUTRAL.

If $pH<7$ , then, clearly, $[H_3O^+]>[HO^-]$ , and the solution is ACIDIC.

If $pH>7$ , then, clearly, $[H_3O^+]<[HO^-]$ , and the solution is ALKALINE.

$8.$ We gots $[H_3O^+]=3.5xx10^-8*mol*L^-1$ .

And given the former.... $pH=-log_10[H_3O^+]=-log_10(3.5xx10^-8)$

$=-(-7.46)=7.46$ , i.e. a basic solution where $[H_3O^+]$ .

$pOH=14-7.46=6.54$ , and $[HO^-]=10^(-6.54)*mol*L^-1=2.85xx10^-7*mol*L^-1$ .

And if we gots $[H_3O^+]=0.0065*mol*L^-1$ , $pH=-log_10(0.0065)=-(-2.19)=2.19$ .

I leave it to you to calculate $pH$ and $pOH$ given the defining relationship..... $14=pH+pOH$

Represent the autoprotolysis of water, and explain how pHpH defines the acidity of the solution. How are pHpH and pOHpOH defined?