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Chapter 4: Problem 113

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

Expert verified
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

01

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
02

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.
03

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.
04

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-}
05

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
06

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%.

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Most popular questions from this chapter

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+& \\ & 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of $6.0 \mathrm{MH}_{2} \mathrm{SO}_{4}\( was spilled, what is the minimum mass of \)\mathrm{NaHCO}_{3}$ that must be added to the spill to neutralize the acid?

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make $1000.0 \mathrm{~L}\( of a \)0.150 \mathrm{M}$ aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

You choose to investigate some of the solubility guidelines for two ions not listed in Table \(4.1,\) the chromate ion \(\left(\mathrm{CrO}_{4}^{2-}\right)\) and the oxalate ion \(\left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right) .\) You are given \(0.01 \mathrm{M}\) solutions \((\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})\) of four water- soluble salts: \begin{tabular}{lll} \hline Solution & Solute & Color of Solution \\ \hline \(\mathrm{A}\) & \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) & Yellow \\ \(\mathrm{B}\) & $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4}$ & Colorless \\ \(\mathrm{C}\) & \(\mathrm{AgNO}_{3}\) & Colorless \\ \(\mathrm{D}\) & \(\mathrm{CaCl}_{2}\) & Colorless \\ \hline \end{tabular} When these solutions are mixed, the following observations are made: \begin{tabular}{lll} \hline Experiment Number & Solutions Mixed & Result \\ \hline 1 & \(\mathrm{~A}+\mathrm{B}\) & Noprecipitate, yellow solution \\ 2 & \(\mathrm{~A}+\mathrm{C}\) & Red precipitate forms \\ 3 & \(\mathrm{~A}+\mathrm{D}\) & Yellow precipitate forms \\ 4 & \(\mathrm{~B}+\mathrm{C}\) & White precipitate forms \\ 5 & \(\mathrm{~B}+\mathrm{D}\) & White precipitate forms \\ 6 & \(\mathrm{C}+\mathrm{D}\) & White precipitate forms \\ \hline \end{tabular} (a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments.

Calculate the concentration of each ion in the following solutions obtained by mixing: (a) \(32.0 \mathrm{~mL}\) of \(0.30 \mathrm{M} \mathrm{KMnO}_{4}\) (b) \(60.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{ZnCl}_{2}^{+}\) with \(15.0 \mathrm{~mL}\) of \(0.60 \mathrm{MKMnO}_{4}\) with \(5.0 \mathrm{~mL}\) of $0.200 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{c}) 4.2 \mathrm{~g}$ of \(\mathrm{CaCl}_{2}\) in \(150.0 \mathrm{~mL}\) of \(0.02 M \mathrm{KCl}\) solution. Assume that the volumes are additive.

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