Question #f4442

Question

Question #f4442

1 Answer

Impact of this question

2235 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-04T11:04:48

pH + pOH = 14

Explanation:

At 25 °C, dissociation constant of water, $K_{w}$ $K_w$ , is equal to $1.0 \times 10^{- 14}$ $1.0×10^{−14}$ ${(\frac{m o l}{L})}^{2}$ $({mol}/L)^2$

$K_{w} = [H_{3} O^{+}] [O H^{-}]$ $K_w=[H_3O^+][OH^-]$
$log (K_{w}) = log ([H_{3} O^{+}] [O H^{-}])$ $log(K_w)=log([H_3O^+][OH^-])$
$log (K_{w}) = log ([H_{3} O^{+}]) + log [O H^{-}])$ $log(K_w)=log([H_3O^+]) + log[OH^-])$
$- 14 = - p H - p O H$ $-14=-pH -pOH$
$14 = p H + p O H$ $14=pH +pOH$

Science

Math

Humanities

... and beyond

Question #f4442

1 Answer

Explanation:

Related questions

Impact of this question