What is the pH of a 0.0067 M KOH solution?

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Answer 1 · 2016-06-03T02:22:37

#pH=11.83#

Potassium hydroxide fully ionizes when dissolved in water according to the following equation:

#KOH -> K^+ + OH^-#

Now, let's find the relationship between #KOH# and #OH^-#

#" " 1 " : "1 " ratio "=> [KOH] =[OH^-]= 6.7xx10^-3M#

In any aqueous solution, # [H_3O^+] #and #[OH^-]# must satisfy the following condition:

#[H_3O^+] [OH^-]= K_w#

#[H_3O^+] = K_w/([OH^-])#

#[H_3O^+] = (1.0xx10^-14)/(6.7xx10^-3)#

#[H_3O^+] = 1.5xx10^-12 M#

Now, after finding the concentration of the hydronium ion, the pH of the solution is determined:

#pH = -log[H_3O^+] #

#pH = - log[1.5xx10^-12]#

#pH=11.83#

Other Method
Find the pOH using the concentration of the hydroxide ion, then use the formula #" "pH + POH = 14 # to find the pH.