# If "0.35 g" of "KHP" neutralized "16 mL" of "NaOH", what is the "NaOH" concentration in molarity?

Apr 3, 2017

$\text{KHP}$, or potassium hydrogen phosphate, is ${\text{K"_2"HPO}}_{4}$, a monoprotic acid. You were given its mass, so you can use its molar mass to find the mols reacted.

To neutralize $\text{NaOH}$, we only need to use one proton, and ${\text{K"_2"HPO}}_{4}$ has one proton. So, the reaction is:

$\text{HPO"_4^(2-)(aq) + "OH"^(-)(aq) -> "PO"_4^(3-)(aq) + "H"_2"O} \left(l\right)$

The molar mass of $\text{KHP}$ is:

$2 \times \text{39.0983 g/mol" + "1.0079 g/mol" + "30.907 g/mol" + 4 xx "15.999 g/mol}$

$=$ $\text{174.1075 g/mol}$

So, the mols used was:

n_("KHP") = "0.35 g KHP" xx "1 mol"/"174.1075 g"

$=$ $\text{0.00201 mols}$

And that is also the $\text{mols}$ of $\text{NaOH}$ because $\text{KHP}$ is monoprotic, and though a weak acid, reacts exactly with the strong base $\text{NaOH}$.

Therefore, from the $\text{mL}$ of $\text{NaOH}$ that were neutralized, the concentration can be found:

["NaOH"] = "0.00201 mols NaOH"/"16 mL" xx "1000 mL"/"L"

$=$ $\text{0.1256 M}$

To two sig figs,

color(blue)(["NaOH"] = "0.13 M")