If #"35.00 mL"# of #"0.211 M"# #"HCl"# was neutralized by #"34.60 mL"# of #"KOH"#, what was the concentration of #"KOH"#?

1 Answer
Truong-Son N.
May 5, 2017

If #"35.00 mL"# of #"0.211 M"# #"HCl"# is neutralized, then that means

#"0.211 M" xx "0.03500 L" = "0.00739 mols H"^(+)#

were neutralized. Since there is one #"OH"^(-)# for every one #"KOH"# and one #"H"^(+)# for every one #"HCl"#, we can say that #"mols H"^(+) = "mols OH"^(-)#. Therefore:

#("0.00739 mols OH"^(-))/("0.03460 L KOH")#

#=# #color(blue)("0.213 M KOH")#.

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