How do you lower your acidity pH?

Question

1619 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-10T18:45:42

Do you not simply add an acid?

In water, we invoke the following equilibrium...

$2H_2O(l) rightleftharpoonsH_3O^+ + HO^-$

And we can write an equilibrium expression, to quantify this dissociation.......

$underbrace(K_w=[HO^-][H_3O^+]=10^-14)_"specified under standard conditions of 298 K and near 1 atm"$

And we take $log_10$ OF BOTH SIDES....

$log_10K_w=log_10[HO^-]+log_10[H_3O^+]=log_(10)10^-14$

But $log_(10)10^-14=-14$

And so $+14=underbrace(-log_10[H_3O^+])_"pH by definition"underbrace(-log_10[HO^-])_"pOH by definition"$

And thus our working relationship... $14=pH+pOH$

And thus if we gots $0.5*mol*L^-1$ $HCl(aq)$ , we gots $[H_3O^+]=0.50*mol*L^-1$ ... $pH=-log_10(0.50)=-(-0.301)=+0.301$ ..

And so $pOH=14-0.301=13.7$ ...

Conc. $HCl(aq)$ is $10.6*mol*L^-1$ out of the bottle. What is $pH$ here?