Question

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

Answer 1

`"KHP"`, or potassium hydrogen phosphate, is `"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 `"NaOH"`, we only need to use one proton, and `"K"_2"HPO"_4` has one proton. So, the reaction is:

`"HPO"_4^(2-)(aq) + "OH"^(-)(aq) -> "PO"_4^(3-)(aq) + "H"_2"O"(l)`

The molar mass of `"KHP"` is:

`2 xx "39.0983 g/mol" + "1.0079 g/mol" + "30.907 g/mol" + 4 xx "15.999 g/mol"`

`=` `"174.1075 g/mol"`

So, the mols used was:

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

`=` `"0.00201 mols"`

And that is also the `"mols"` of `"NaOH"` because `"KHP"` is monoprotic, and though a weak acid, reacts exactly with the strong base `"NaOH"`.

Therefore, from the `"mL"` of `"NaOH"` that were neutralized, the concentration can be found:

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

`=` `"0.1256 M"`

To two sig figs,

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