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

May 5, 2017

If $\text{35.00 mL}$ of $\text{0.211 M}$ $\text{HCl}$ is neutralized, then that means

${\text{0.211 M" xx "0.03500 L" = "0.00739 mols H}}^{+}$

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

$\left(\text{0.00739 mols OH"^(-))/("0.03460 L KOH}\right)$

$=$ $\textcolor{b l u e}{\text{0.213 M KOH}}$.