Acid base Equilibrium?

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1 Answer
Truong-Son N.
Apr 4, 2018

Well, we assume #"HCl"# is an ideal gas, so that

#PV = nRT#

where #P#, #V#, #n#, #R#, and #T# are the pressure in #"atm"#, volume in #"L"#, mols (in #"mols"#, of course!), universal gas constant in #"L"cdot"atm/mol"cdot"K"#, and the temperature in #"K"#, respectively.

So the mols of #"HCl"# at STP (before 1982) are:

#n = (PV)/(RT) = ("1 atm" cdot "0.241 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "273.15 K")#

#=# #"0.0108 mols"#

within #"2.0 L"# of water, giving #"0.00538 M"# #"HCl"(aq)#. #"HCl"# is well-known to be a strong acid, so it dissociates completely (especially at this low concentration!) to give #"0.00538 M H"^(+)#.

Therefore

#color(blue)("pH") = -log["H"^(+)] = color(blue)(2.27)#

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