Problem 113 - Chemistry: The Central Science in SI Units, Global Edition

Question

The arsenic in a \(1.22-g\) sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of As in \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-} ?(\mathbf{b})\) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Step-by-Step Solution

Answer

The oxidation state of As in AsO\(_{4}^{3-}\) is +5. The compound Ag\(_{3}\)AsO\(_{4}\) is named Silver arsenate. The mass percentage of arsenic in the pesticide is 5.22%.

Step 1: Determine the oxidation state of As in AsO4^{3-}

To find the oxidation state of As in AsO4^{3-}, consider the following rules: 1. The oxidation state of oxygen is usually -2. 2. The sum of the oxidation states in a compound must equal the charge on the compound. Let x represent the oxidation state of As. In AsO4^{3-}, there are 4 oxygen atoms, each with a -2 oxidation state. So, the equation is: x + 4(-2) = -3

Step 2: Solve for the oxidation state of As

Now, solve the equation from Step 1: x - 8 = -3 x = -3 + 8 x = 5 The oxidation state of As in AsO4^{3-} is +5.

Step 3: Name Ag3AsO4 by analogy to the corresponding phosphorus-containing compound

The corresponding phosphorus-containing compound is Ag3PO4. This compound is called Silver phosphate. Replacing the phosphorus atom with an arsenic atom, we get Ag3AsO4. Therefore, the name of Ag3AsO4 is Silver arsenate.

Step 4: Calculate the moles of AsO4^{3-} using the titration data

We are given the volume (25.0 mL) and molarity (0.102 M) of the Ag^+ solution. At the equivalence point in the titration, the moles of Ag^+ will equal three times the moles of AsO4^{3-} as per the following balanced reaction: 3 Ag^+ + AsO4^{3-} -> Ag3AsO4 moles of Ag^+ = (volume in L) × (molarity) moles of Ag^+ = (25.0 mL × \(\frac{1 L}{1000 mL}\)) × 0.102 M = 0.00255 mol Let y represent the moles of AsO4^{3-}. Then, according to the balanced reaction: 3y = 0.00255 mol y = 0.00255/3 mol y = 0.00085 mol AsO4^{3-}

Step 5: Calculate the mass of arsenic (As) in the sample

Now, we need to calculate the mass of arsenic in the sample. We know there are 0.00085 mol AsO4^{3-}. Since there is one As atom per AsO4^{3-} unit, we also have 0.00085 mol As. mass of As = (moles of As) × (molar mass of As) mass of As = 0.00085 mol × 74.92 g/mol = 0.06368 g

Step 6: Calculate the mass percentage of arsenic in the pesticide

Now, to find the mass percentage of arsenic in the pesticide, we will use the following formula: mass percentage = \(\frac{mass~of~As}{mass~of~pesticide~sample}\) × 100% mass percentage = \(\frac{0.06368~g}{1.22~g}\) × 100% = 5.22% The mass percentage of arsenic in the pesticide is 5.22%.

Key Concepts

Oxidation State

Understanding oxidation state is key to figuring out how elements behave in compounds. Let's start with arsenic in the arsenate ion, \(AsO_4^{3-}\). The oxidation state tells us the number of electrons an atom gains or loses when forming a compound. Here, each oxygen atom typically has an oxidation state of -2. Multiply this by the four oxygens in the compound, and you get -8 for oxygen alone. The overall charge of the ion is -3.
Now, let's balance the equation: we know the total oxidation states must equal the ion's charge. Let the oxidation state of arsenic be \(x\). The equation becomes \(x + 4(-2) = -3\). Solving this gives \(x = +5\). So, arsenic has an oxidation state of +5 in \(AsO_4^{3-}\). This understanding is crucial because it helps chemists determine how substances will react in different situations.

Silver Arsenate

Silver arsenate, known chemically as \(Ag_3AsO_4\), plays a key role in titration processes. To name it, we compare it to a more familiar compound, silver phosphate \(Ag_3PO_4\). By swapping the phosphorus in silver phosphate for arsenic, we derive the name 'Silver Arsenate.' This approach is common in chemistry; it helps in identifying compounds with different central atoms but similar structures.
This compound is pivotal during titration because it forms as a precipitate as the \(Ag^+\) ions bind with \(AsO_4^{3-}\) ions. Understanding how compounds are named helps in identifying them quickly in chemical reactions and predicting reaction outcomes efficiently.

Mass Percentage Calculation

Mass percentage gives a clear picture of how much of a substance is in a given sample, in this case, arsenic in a pesticide. Calculating this involves several steps, often starting with a titration to find the amount of a reactant needed to reach the equivalence point.
In the exercise, the equivalent point was achieved with \(25.0 \, \text{mL}\) of a \(0.102 \, M\) \(Ag^+\) solution. This tells us about the moles of \(Ag^+\) used, which must be three times the moles of \(AsO_4^{3-}\) due to the stoichiometry of \(3 \, \text{Ag}^+ + \text{AsO}_4^{3-} \rightarrow \text{Ag}_3\text{AsO}_4\). From the titration data, \(0.00255 \, \text{mol}\) of \(Ag^+\) corresponds with \(0.00085 \, \text{mol}\) of arsenate.
Finally, multiply the moles of arsenic by its molar mass (\(74.92 \, \text{g/mol}\)) to find its mass, 0.06368 g, in the pesticide. To find mass percentage, divide the mass of arsenic by the sample mass, then multiply by 100. This gives a 5.22% arsenic concentration in the pesticide. This calculation is a critical tool in quantifying substances in complex mixtures.