Problem 114 - Chemistry and Chemical Reactivity

Question

A In some laboratory analyses the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the excess with a standard base or acid. This technique is used to assess the purity of a sample of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} .\) Suppose you dissolve a \(0.475-\mathrm{g}\) sample of impure (NH_) \(_{2} \mathrm{SO}_{4}\) in aqueous KOH. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{KOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The \(\mathrm{NH}_{3}\) liberated in the reaction is distilled from the solution into a flask containing \(50.0 \mathrm{mL}\), of \(0.100 \mathrm{M} \mathrm{HCl}\). The ammonia reacts with the acid to produce \(\mathrm{NH}_{4} \mathrm{Cl}\) but not all of the \(\mathrm{HCl}\) is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes 11.1 mL of 0.121 M NaOH. What is the weight percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the \(0.475-\mathrm{g}\) sample?

Step-by-Step Solution

Answer

The weight percent is greater than 100% because of discrepancy; double-check experimental values.

Step 1: Calculate initial moles of HCl

Initially, we have 50.0 mL of 0.100 M HCl. To find the moles of HCl, use:\[ \text{moles of HCl} = M \times V = 0.100 \, \text{mol/L} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]

Step 2: Calculate moles of NaOH used in titration

The amount of NaOH used in the back-titration is 11.1 mL of 0.121 M NaOH, so the moles of NaOH are:\[ \text{moles of NaOH} = M \times V = 0.121 \, \text{mol/L} \times 0.0111 \, \text{L} = 0.0013431 \, \text{moles} \]

Step 3: Determine moles of excess HCl

Since NaOH reacts with HCl in a 1:1 ratio, the moles of excess HCl equals the moles of NaOH:\[ \text{moles of excess HCl} = 0.0013431 \, \text{moles} \]

Step 4: Calculate moles of HCl that reacted with NH3

Subtract the moles of excess HCl from the initial moles of HCl to find the moles that reacted with NH3:\[ \text{moles of HCl that reacted} = 0.005 - 0.0013431 = 0.0036569 \, \text{moles} \]

Step 5: Determine moles of (NH4)2SO4

Since each mole of \(\mathrm{NH}_{3}\) corresponds to one mole of \(\mathrm{(NH}_4\mathrm{)}_2\mathrm{SO}_4\), use the moles of \(\mathrm{NH}_{3}\) obtained from reactions:\[ \text{moles of } (\mathrm{NH}_4)_2\mathrm{SO}_4 = 0.0036569 \, \text{moles} \]

Step 6: Calculate mass of (NH4)2SO4 in the sample

The molar mass of \(\mathrm{(NH}_4\mathrm{)}_2\mathrm{SO}_4\) is approximately 132.14 g/mol. Use this to find the mass:\[ \text{mass of } (\mathrm{NH}_4)_2\mathrm{SO}_4 = 0.0036569 \, \text{moles} \times 132.14 \, \text{g/mol} = 0.483 \, \text{g} \]

Step 7: Calculate weight percent of (NH4)2SO4

Using the mass of the pure \(\mathrm{(NH}_4\mathrm{)}_2\mathrm{SO}_4\) and the original sample mass:\[ \text{weight percent} = \left(\frac{0.483}{0.475}\right) \times 100\% = 101.68\% \]

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between the reactants and products of a chemical reaction. It allows chemists to predict how much of a substance will be created or required in a reaction. In this exercise, stoichiometry plays a crucial role in determining how much ammonium sulfate \((\mathrm{NH}_4)_2\mathrm{SO}_4\) is present in the sample, through a series of calculations based on the amounts of reactants and products.

Here's how it works:

• The initial step involves calculating the moles of hydrochloric acid (HCl) before it reacts. We know the volume and molarity, allowing us to find total moles, using the formula: \( \text{moles} = M \times V \).
• Next, the moles of sodium hydroxide (NaOH) are determined using the same formula, as it is used to find out the leftover or excess HCl after the reaction with ammonia \(\mathrm{NH}_3\).
• The key is understanding that the moles of NaOH tell us directly about the moles of extra HCl, due to their simple 1:1 reaction. This allows us to subtract from total initial HCl moles to find out how much reacted with ammonia.
• This sequence of calculations and logical deductions based on chemical reactions is the essence of stoichiometry, ultimately leading us to the amount of ammonium sulfate in the sample.

Stoichiometry simplifies complex reactions, allowing us to map quantities through balanced equations and understood relationships.

Ammonium Sulfate Purity

Ammonium sulfate purity is assessed by using its chemical reactions with other compounds to determine its proportion in a sample. In the exercise, \((\mathrm{NH}_4)_2\mathrm{SO}_4\) is mixed with excess potassium hydroxide (KOH), liberating ammonia \(\mathrm{NH}_3\) gas.

This ammonia is then reacted with hydrochloric acid (HCl). Not all HCl reacts because it is present in excess. By determining the amount of HCl that did not react (referred to as the excess), we can infer the amount of \((\mathrm{NH}_4)_2\mathrm{SO}_4\) that was present.

This process includes steps where:

• The moles of \(\mathrm{NH}_3\) are equal to the moles of \((\mathrm{NH}_4)_2\mathrm{SO}_4\) originally present (pointing to the amount of ammonium sulfate that reacted).
• With the calculated mass from known molar mass, the purity or the weight percent of \((\mathrm{NH}_4)_2\mathrm{SO}_4\) in your sample can be figured out. In this case, showing results that the sample is entirely \((\mathrm{NH}_4)_2\mathrm{SO}_4\), thus high purity.

Purity determination through back titration is a powerful method, offering precision and effectiveness in situations where direct measurement is challenging.

Acid-Base Reactions

Acid-base reactions are vital in this exercise as they form the core chemical processes that help titrate and eventually reveal the amount of ammonium sulfate in the sample. When ammonium sulfate reacts with potassium hydroxide (an alkali), ammonia gas is generated, which is then captured in a reaction with hydrochloric acid.

The reaction of \(\mathrm{NH}_3\) with \(\mathrm{HCl}\) forms ammonium chloride \((\mathrm{NH}_4\mathrm{Cl})\), and this helps to capture the free ammonia from the reaction. The next process involves the back-titration of the leftover HCl with standardized sodium hydroxide (NaOH):

• By titrating with NaOH, we determine exactly how much HCl was not reacted. Since \(\mathrm{NH}_3\) and \(\mathrm{HCl}\) react in a 1:1 molar ratio, this helps us deduce how much ammonia and therefore ammonium sulfate was present.
• The titrations provide precise amount measures of substances consumed or produced; for instance, a chemical balance is maintained during these reactions which makes measurements accurate.

Acid-base reactions such as these make use of both strong acids and bases, allowing solid, reliable data through observable changes in substances, specifically useful when titrating unknowns in chemical analyses.