A student added 5.00 g of \({{\text{P}}_{\text{4}}}{{\text{O}}_{{\text{10}}}}\) to 1.50 g of water. Determine the limiting reactant, showing your working.

Calculate the mass of phosphoric(V) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\), formed in the reaction.

Phosphoric(V) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}\), has a \({\text{p}}{K_{\text{a}}}\) of 2.12 (\({\text{p}}{K_{{\text{a1}}}}\)) while phosphoric(III) acid, \({{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}\), has a \({\text{p}}{K_{\text{a}}}\) of 1.23 (\({\text{p}}{K_{{\text{a1}}}}\)). Identify the weaker of the two acids, giving a reason for your choice.

\({{\text{P}}_4}{{\text{O}}_{10}}{\text{: }}\left( {\frac{{{\text{5.00}}}}{{{\text{283.88}}}} = } \right){\text{ 0.0176 (mol)}}\) and \({{\text{H}}_2}{\text{O: }}\left( {\frac{{{\text{1.50}}}}{{{\text{18.02}}}} = } \right){\text{ 0.0832 (mol)}}\);

\({{\text{H}}_2}{\text{O}}\) is the limiting reactant and reason related to stoichiometry;

\(\frac{{0.0832 \times 4}}{6}/0.0555{\text{ (mol)}}\);

\((0.0555 \times 98.00 = ){\text{ }}5.44{\text{ g}}\);

The unit is needed for M2.

Award [2] for correct final answer.

Do not penalize slight numerical variations due to premature rounding.

\({{\text{H}}_3}{\text{P}}{{\text{O}}_4}\) is the weaker acid and higher \({\text{p}}{K_{\text{a}}}\)/lower \({K_{\text{a}}}\);

The majority of candidates calculated the amounts of reactants correctly, and many of them applied the stoichiometric ratio correctly to determine the limiting reactant.

More than half of the candidates calculated the mass of product correctly. Even if the final result was incorrect quite frequently students gained some credit through the application of ECF.

Many candidates appreciated that a higher \({\text{p}}{K_{\text{a}}}\) means a weaker acid. Some candidates did not refer to the \({\text{p}}{K_{\text{a}}}\) or \({K_{\text{a}}}\) value in their reasoning, failing to score a mark.