Question

Ten \(\mathrm{mL}\) of concentrated \(\mathrm{H}_{3} \mathrm{PO}_{4} \quad(91.7 \%\) by mass, \(d=1.69 \mathrm{~g} / \mathrm{mL})\) was accidentally poured into a beaker containing \(20.0 \mathrm{~g}\) of \(\mathrm{NaOH} .\) Not all the \(\mathrm{NaOH}\) was consumed. How many grams of \(\mathrm{NaOH}\) were left unreacted? The equation for the reaction is $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{OH}^{-}(a q) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}+\mathrm{PO}_{4}^{3-}(a q) $$

Short answer

Answer: The mass of unreacted NaOH after the reaction is 0.896 grams.

Step-by-step solution

Find the mass of H3PO4

Given volume and density of H3PO4, we can find the mass by using the equation mass = volume × density: mass of H3PO4 = 10 mL × 1.69 g/mL = 16.9 g

Find the mass of pure H3PO4 in the concentrated solution

Since the concentration of H3PO4 is given as 91.7% by mass, the mass of pure H3PO4 can be calculated as follows: mass of pure H3PO4 = (91.7 / 100) × 16.9 g = 15.4943 g

Calculate the moles of H3PO4 and OH- ions

To find the moles of H3PO4 and OH-, divide the mass by the respective molar mass: moles of H3PO4 = 15.4943 g / (3×1.01 + 15.999 + 4×15.999) g/mol = 0.159 mol moles of OH- = 20.0 g / (22.99 + 15.999) g/mol = 0.500 mol

Determine the moles of OH- ions that react with H3PO4

Using the balanced chemical equation, we find that 3 moles of OH- react with 1 mole of H3PO4. Therefore, the moles of OH- that react with H3PO4 can be calculated by: moles of reacted OH- = 3 × moles of H3PO4 = 3 × 0.159 mol = 0.477 mol

Calculate the moles of unreacted OH- ions

Since the moles of reacted OH- ions were determined in step 4, we can find the moles of unreacted OH- ions by subtracting reacted OH- ions from the initial moles of OH- ions: moles of unreacted OH- = moles of OH- - moles of reacted OH- = 0.500 mol - 0.477 mol = 0.023 mol

Calculate the mass of unreacted NaOH

To find the mass of unreacted NaOH, multiply the moles of unreacted OH- ions by the molar mass of NaOH: mass of unreacted NaOH = 0.023 mol × (22.99 + 15.999) g/mol = 0.896 g Therefore, after the reaction, 0.896 grams of NaOH are left unreacted.

Key concepts

Mole Concept

Understanding the mole concept is crucial in solving stoichiometry problems. Essentially, a mole is a unit in chemistry that denotes a specific number, much like a dozen means twelve. However, a mole, abbreviated as 'mol', is equal to Avogadro's number, which is approximately 6.022 × 10^23. This is the number of atoms in 12 grams of carbon-12, and it's the same number of units, be it atoms, molecules, or ions, in any substance.

When faced with chemical reactions, we need to calculate the amount of substances involved. Instead of counting individual atoms or molecules, chemists use the mole concept to group particles into manageable units—moles. To convert the mass of a substance into moles, we divide by its molar mass - the mass of one mole of that substance, which is numerically equal to the substance's atomic or molecular weight in grams.

For example, the molar mass of water (H2O) is about 18.015 g/mol, meaning one mole of water has a mass of 18.015 grams. In stoichiometry problems, it's necessary to convert mass to moles to use the balanced chemical equation, which tells us how moles of different substances interact. Thus, the mole concept is a bridge between the mass of substances we measure in the lab and the proportional relationships in a chemical reaction.

Chemical Reaction

A chemical reaction is a process where substances, called reactants, are transformed into new substances, called products. The changes involve the breaking and forming of chemical bonds, leading to a rearrangement of atoms. Chemical reactions are represented by chemical equations, which are symbolic representations where the reactants are written on the left side and the products on the right, separated by an arrow indicating the direction of the reaction.

In the chemical equation for the reaction between phosphoric acid (H3PO4) and hydroxide ions (OH-), written as:
\[\mathrm{H}_{3} \mathrm{PO}_{4}(aq) + 3 \mathrm{OH}^{-}(aq) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O} + \mathrm{PO}_{4}^{3-}(aq)\]
Each substance has a coefficient that represents the number of moles involved in the reaction. For example, the '3' before \(\mathrm{OH}^{-}\) indicates that three moles of hydroxide ions react with one mole of phosphoric acid. Understanding the coefficients is vital as they are used to establish the ratios that are necessary to determine the amount of products formed or reactants consumed in the reaction.

Limiting Reactant

The concept of the limiting reactant is fundamental in stoichiometry and refers to the substance that is entirely consumed first during a chemical reaction, thus determining the amount of product that can be formed. Since reactants do not always mix in the exact ratios indicated by the balanced chemical equation, one reactant will limit the extent of the reaction.

Identifying the limiting reactant is a two-step process: first, we calculate the moles of each reactant, then we use the stoichiometric ratios from the balanced chemical equation to determine which reactant will run out first. That reactant is the limiting one. Upon recognition of the limiting reactant, we can predict how much product will be formed and how much of the excess reactant will remain unreacted.

In our given problem, the reaction between H3PO4 and NaOH, had excess NaOH left over, indicating that H3PO4 was the limiting reactant. After all the H3PO4 has reacted, there will be some NaOH which did not get to react - this is the unreacted NaOH. The calculation of the unreacted NaOH requires finding the difference between the initial moles of NaOH and the moles that reacted with the available H3PO4. Understanding the limiting reactant concept not only helps in predicting the outcome of reactions but also in optimizing the use of reactants in industrial and laboratory settings.