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

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

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Answer 1 · 2017-04-03T06:59:34

$KHP$ $"KHP"$ , or potassium hydrogen phosphate, is $K_{2} {HPO}_{4}$ $"K"_2"HPO"_4$ , a monoprotic acid. You were given its mass, so you can use its molar mass to find the mols reacted.

To neutralize $NaOH$ $"NaOH"$ , we only need to use one proton, and $K_{2} {HPO}_{4}$ $"K"_2"HPO"_4$ has one proton. So, the reaction is:

${HPO}_{4}^{2 -} (a q) + {OH}^{-} (a q) \to {PO}_{4}^{3 -} (a q) + H_{2} O (l)$ $"HPO"_4^(2-)(aq) + "OH"^(-)(aq) -> "PO"_4^(3-)(aq) + "H"_2"O"(l)$

The molar mass of $KHP$ $"KHP"$ is:

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

$=$ $=$ $174.1075 g/mol$ $"174.1075 g/mol"$

So, the mols used was:

$n_{KHP} = 0.35 g KHP \times \frac{1 mol}{174.1075 g}$ $n_("KHP") = "0.35 g KHP" xx "1 mol"/"174.1075 g"$

$=$ $=$ $0.00201 mols$ $"0.00201 mols"$

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

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

$[NaOH] = \frac{0.00201 mols NaOH}{16 mL} \times \frac{1000 mL}{L}$ $["NaOH"] = "0.00201 mols NaOH"/"16 mL" xx "1000 mL"/"L"$

$=$ $=$ $0.1256 M$ $"0.1256 M"$

To two sig figs,

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

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

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