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

Question

Wyatt Atkins · Accepted Answer

#"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"#

Thus, the mols employed were:

#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 significance figures,

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

Olivia Anton · Answer

undefined To find the concentration of NaOH in molarity, use the formula:

$$ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $$

First, calculate the moles of KHP:

$$ \text{moles of KHP} = \frac{\text{mass of KHP (g)}}{\text{molar mass of KHP (g/mol)}} $$

Then, use the stoichiometry of the reaction to find moles of NaOH:

$$ \text{moles of NaOH} = \text{moles of KHP} \times \frac{\text{mol of NaOH}}{\text{mol of KHP}} $$

Finally, divide moles of NaOH by the volume of solution in liters to get the molarity of NaOH.