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 \(\mathrm{AsO}_{4}{ }^{3-} ?(\mathbf{b})\) 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?

Short 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-by-step solution

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

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.

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.

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-}

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

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.