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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