How do you lower your acidity pH?

1 Answer
Jul 10, 2018

Do you not simply add an acid?

Explanation:

In water, we invoke the following equilibrium...

$2 {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

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

${\underbrace{{K}_{w} = \left[H {O}^{-}\right] \left[{H}_{3} {O}^{+}\right] = {10}^{-} 14}}_{\text{specified under standard conditions of 298 K and near 1 atm}}$

And we take ${\log}_{10}$ OF BOTH SIDES....

${\log}_{10} {K}_{w} = {\log}_{10} \left[H {O}^{-}\right] + {\log}_{10} \left[{H}_{3} {O}^{+}\right] = {\log}_{10} {10}^{-} 14$

But ${\log}_{10} {10}^{-} 14 = - 14$

And so $+ 14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{\text{pH by definition"underbrace(-log_10[HO^-])_"pOH by definition}}$

And thus our working relationship...$14 = p H + p O H$

And thus if we gots $0.5 \cdot m o l \cdot {L}^{-} 1$ $H C l \left(a q\right)$, we gots $\left[{H}_{3} {O}^{+}\right] = 0.50 \cdot m o l \cdot {L}^{-} 1$...$p H = - {\log}_{10} \left(0.50\right) = - \left(- 0.301\right) = + 0.301$..

And so $p O H = 14 - 0.301 = 13.7$...

Conc. $H C l \left(a q\right)$ is $10.6 \cdot m o l \cdot {L}^{-} 1$ out of the bottle. What is $p H$ here?