One litre of an aqueous solution of KOH has a pH of 12 at 25 celcius degrees. (continue below)The amount of pure HCl gas, in mol, that must be added to the solution to lower the pH from 12 to 2 is...

One litre of an aqueous solution of KOH has a pH of 12 at 25 celcius degrees.

(continue below)

The amount of pure HCl gas, in mol, that must be added to the solution to lower the pH from 12 to 2 is 0.02.

explain please.

Expert Answers
nessus eNotes educator| Certified Educator

1st calculate concentration of KOH solution:

pH 12 means [H+](concentration H+) = 10^-12.  Use this value in equilibrium expression for water to find [OH-].

[H+][OH-] = 10^-14,  [OH-] = 10^-14/[H+] = 10^-14/10^-12 = 10^-2 (0.01) moles per litre OH-

Concentration, c(KOH)  = c(OH-)  =  0.01moles per litre.

From  HCl  +  KOH  =  KCl  +  H2O  adding 0.02 moles HCl to  0.01 moles KOH means 0.01(10^-2) moles of HCl remaining in solution.

New pH =  -log[H+]  =  -log 10^-2  = 2

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