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When we talk about percent purity in chemistry, we're measuring the amount of a specified pure substance in a sample, expressed as a percentage. In this exercise involving \(\mathrm{KH}_{2}\mathrm{PO}_{4}\), percent purity helps us understand how much of the sample is actually the pure compound versus impurities or other substances.

To calculate percent purity:

• Find the theoretical mass of the pure compound that should react based on the experimental conditions.
• Compare this to the actual mass used in the experiment.
• Use the formula for percent purity: \[\text{percent purity} = \left(\frac{\text{mass of pure compound}}{\text{mass of sample}}\right) \times 100\]

In the given problem, we calculated that the theoretical mass of pure \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) is 0.39022 g from the titration. The actual sample mass used was 0.492 g. Hence, the percent purity was determined to be approximately 79.31%.

Molar mass is a fundamental concept that refers to the mass of one mole of a given substance. It's the sum of the atomic masses of all atoms present in a molecule, most often expressed in grams per mole (g/mol).

To compute the molar mass of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\), you add the atomic masses of each element:

• Potassium (K): approximately 39.10 g/mol
• Hydrogen (H): approximately 1.01 g/mol, with two hydrogen atoms totaling 2.02 g/mol
• Phosphorus (P): approximately 30.97 g/mol
• Oxygen (O): approximately 16.00 g/mol, with four oxygen atoms totaling 64.00 g/mol

These sum up to give the molar mass of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) as about 136.09 g/mol. This molar mass is crucial for converting between the mass of the substance and the number of moles, as seen in titration and other chemical calculations.

Chemical equations are expressions that describe the chemical transformations occurring in a reaction. They provide vital details about the reactants and products, as well as their respective coefficients which indicate the proportionate amounts involved.

For the titration exercise discussed, the balanced chemical equation is:\[\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{OH}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}_{2}\mathrm{O}\]This equation shows that one mole of dihydrogen phosphate ion \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) reacts with one mole of hydroxide ion \(\mathrm{OH}^{-}\) to yield one mole of hydrogen phosphate ion \(\mathrm{HPO}_{4}^{2-}\) and water \(\mathrm{H}_{2}\mathrm{O}\).

Understanding this reaction is crucial because it tells us that the titration uses exactly equal molar amounts of acid and base, allowing us to determine the purity of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) accurately based on the volume and molarity of the titrant used.

Analytical calculation in chemistry involves using measurements and calculations to determine an unknown quantity. In this exercise, it's all about determining the percent purity of a chemical sample through titration.

The process follows these steps:

• Calculate the moles of titrant (here being \(\mathrm{NaOH}\)) used, using the formula: \[\text{moles} = \text{concentration} \times \text{volume}\]This helps determine how much \(\mathrm{OH}^-\) is needed to react completely with \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\).
• Use the balanced chemical equation to find the moles of \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) which is equivalent to the moles of \(\mathrm{NaOH}\) used.
• Calculate the theoretical mass of pure \(\mathrm{KH}_{2}\mathrm{PO}_{4}\) by multiplying the moles obtained with its molar mass.
• Compare this theoretical mass to the actual sample mass to find the percent purity.

Through such calculations, chemists can analyze the composition and quality of chemical substances with precision.