What is the "pH" for an aqueous solution with a hydroxide concentration of 5 xx 10^(-3) "M" at 25^@ "C"?

2 Answers
Dec 25, 2017

11.7

Explanation:

"pOH" = - log["OH"^-]

"pOH" = - log(5 × 10^-3) = 3 - log5 = 2.30

Now,

"pH" + "pOH" = 14

"pH" = 14 - "pOH" = 14 - 2.30 = 11.7

"pH" ~~ color(blue)(11.7)

Explanation:

color (indigo)(pH=-log [H_3O^+])
color (indigo)(pOH=-log [OH^-])
color(cyan)([OH^-]=0.005)
Here,
pOH=-log [0.005]
pOH=-log [5×10^-3]
pOH=2.3010299957

color (green)(pH=14-pOH)

ph=14-2.301 =11.699 ~~ bb11.7