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The concentration of cyanide, \(\mathrm{CN}^{-},\) in a copper electroplating bath is determined by a complexometric titration using \(\mathrm{Ag}^{+}\) as the titrant, forming the soluble \(\mathrm{Ag}(\mathrm{CN})_{2}^{-}\) complex. In a typical analysis a \(5.00-\mathrm{mL}\) sample from an electroplating bath is transferred to a 250 -mL Erlenmeyer flask, and treated with \(100 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}, 5 \mathrm{~mL}\) of \(20 \% \mathrm{w} / \mathrm{v}\) \(\mathrm{NaOH}\) and \(5 \mathrm{~mL}\) of \(10 \% \mathrm{w} / \mathrm{v}\) KI. The sample is titrated with 0.1012 \(\mathrm{M} \mathrm{AgNO}_{3}\), requiring \(27.36 \mathrm{~mL}\) to reach the end point as signaled by the formation of a yellow precipitate of AgI. Report the concentration of cyanide as parts per million of \(\mathrm{NaCN}\).

Short Answer

Expert verified
The concentration of cyanide as parts per million of \( \mathrm{NaCN} \) is 27146 ppm.

Step by step solution

01

Determine Moles of AgNO3 Used

Calculate the moles of AgNO3 used in the titration. Use the formula \( ext{moles} = ext{concentration} \times ext{volume} \). The concentration of AgNO3 is given as 0.1012 M, and the volume used is 27.36 mL, which needs to be converted to liters: 0.02736 L. Therefore, \( ext{moles of } \mathrm{AgNO_{3}} = 0.1012 \, \text{M} \times 0.02736 \, \text{L} = 0.00277 \, \text{moles} \).
02

Relate Moles of AgNO3 to Moles of CN-

The reaction between \( \mathrm{Ag}^{+} \) and \( \mathrm{CN}^{-} \) is assumed to create a 1:1 molar relationship because each \( \mathrm{Ag}^{+} \) combines with two \( \mathrm{CN}^{-} \) ions, forming \( \mathrm{Ag(CN)_2^{-}} \). Hence, the moles of \( \mathrm{CN}^{-} \) can be considered equal to the moles of \( \mathrm{AgNO}_{3} \), which is 0.00277 moles.
03

Convert Moles of CN- to Mass of NaCN

To convert moles of \( \mathrm{CN}^{-} \) to mass of \( \mathrm{NaCN} \), use the molar mass of \( \mathrm{NaCN} \), which is approximately 49 g/mol. Thus, the mass of \( \mathrm{NaCN} \) is \( 0.00277 \, \text{moles} \times 49 \, \text{g/mol} \approx 0.13573 \, \text{g} \).
04

Determine Concentration in Parts Per Million (ppm)

Parts per million (ppm) is calculated by \( \text{ppm} = \left( \frac{\text{mass of solute}}{\text{volume of solution (in L)}} \right) \times 10^6 \). The volume of the sample is 5.00 mL or 0.005 L. So, \( \text{ppm} = \left( \frac{0.13573 \, \text{g}}{0.005 \, \text{L}} \right) \times 10^6 = 27146 \, \text{ppm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cyanide Concentration Determination
Determining the concentration of cyanide ions in a solution is critical for safety and effectiveness in industrial applications. This involves a method called complexometric titration.
The primary aim is to find out how much cyanide (- \( \mathrm{CN}^{-} \)) is present in a solution, such as a copper electroplating bath.
In this process, we use a titrant, \( \mathrm{AgNO}_3 \), which reacts specifically with cyanide ions to form a soluble complex, \( \mathrm{Ag(CN)_2^{-}} \).
  • First, a known volume of the sample is prepared in an Erlenmeyer flask with the addition of water and specific reagents like \( \mathrm{NaOH} \) and \( \mathrm{KI} \) to set the right conditions for the titration to occur effectively.
  • The titration continues until a distinct yellow precipitate of \( \mathrm{AgI} \) is observed, signaling the endpoint.
  • This method gives an accurate measure of how much cyanide is present, allowing calculations converted to concentration units like parts per million (ppm).
Understanding how to calculate and report these values is essential for compliance with industrial and safety regulations.
Copper Electroplating Bath
Copper electroplating is a process in which a layer of copper is deposited onto a substrate material. This technique uses an electroplating bath containing a solution with copper ions and other chemicals, like cyanide, to facilitate the plating process.
  • The bath's composition must be precisely monitored and controlled for effective plating results and safe handling.
  • Cyanide is often included in these solutions due to its ability to form stable complexes with metals, aiding even deposition of copper on surfaces.
  • Regular analysis, including determining cyanide concentration, is necessary to ensure the bath remains at optimal conditions and prolongs its efficiency.
This regulation is crucial as excessive cyanide concentrations can pose safety hazards and negatively affect the quality of the electroplating process.
Analytical Chemistry
Analytical chemistry focuses on the composition of material substances. In the context of this exercise, it plays a pivotal role in determining the concentration of specific ions—like cyanide—in complex mixtures such as electroplating baths.
  • The methods used, such as complexometric titration, are designed to achieve precise and accurate results.
  • This field involves both qualitative and quantitative analysis, providing insights into the chemical makeup and concentrations.
  • The rigorous training in method development, validation, and implementation ensures reliability in measurements.
In industrial settings, such precise analysis supports maintaining product quality and safety standards.
Molar Calculations
Molar calculations are essential for determining the concentration of substances in chemical solutions. These calculations involve concepts such as molarity, which is defined as moles of solute per liter of solution.
  • In titrations, you calculate the number of moles of the titrant based on its molarity and the volume used in the reaction.
  • A typical equation used is: \[\text{moles} = \text{concentration (M)} \times \text{volume (L)}\]
  • These moles can then be related to those of the substance being measured, using the balanced chemical equation for the reaction.
  • Accurate molar calculations are foundational in converting results into meaningful data, such as ppm or mass concentration.
By mastering these calculations, you improve your ability to analyze and predict the behavior of chemical solutions.
Parts Per Million (ppm)
Parts per million (ppm) is a unit of concentration often used to express very dilute concentrations of substances. This measure is essential for understanding the presence of a substance in a mixture and ensuring it meets safety standards.
  • Ppm is calculated as:\[\text{ppm} = \left( \frac{\text{mass of solute}}{\text{volume of solution (in L)}} \right) \times 10^6\]
  • This calculation provides a clear picture of how much of a substance exists within a larger volume.
  • Converting concentrations into ppm is particularly useful in industrial settings and environmental science to gauge pollutant levels.
Using ppm, industries can ensure compliance with regulations and maintain safe operation levels.
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are chemical processes where electrons are transferred between substances. These reactions are vital in many industrial applications, including electroplating baths.
  • In the context of copper electroplating, oxidation and reduction help deposit copper onto desired surfaces.
  • The cyanide assists in forming complexes with metals, hence facilitating these redox reactions efficiently.
  • Understanding and controlling these reactions are essential to achieving precise electroplating outcomes.
Mastery of redox reactions helps in managing chemical processes safely and effectively, ensuring desired results in production.

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Most popular questions from this chapter

The concentration of \(o\) -phthalic acid in an organic solvent, such as \(n\) butanol, is determined by an acid-base titration using aqueous \(\mathrm{NaOH}\) as the titrant. As the titrant is added, the \(o\) -phthalic acid extracts into the aqueous solution where it reacts with the titrant. The titrant is added slowly to allow sufficient time for the extraction to take place. (a) What type of error do you expect if the titration is carried out too quickly? (b) Propose an alternative acid-base titrimetric method that allows for a more rapid determination of the concentration of \(o\) -phthalic acid in \(n\) -butanol.

Calculate or sketch titration curves for the following acid-base titrations. a. \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.0500 \mathrm{M} \mathrm{HCl}\) b. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCOOH}\) with \(0.100 \mathrm{M} \mathrm{NaOH}\) c. \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\) with \(0.100 \mathrm{M} \mathrm{HCl}\) d. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M}\) ethylenediamine with \(0.100 \mathrm{M} \mathrm{HCl}\) e. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M}\) citric acid with \(0.120 \mathrm{M} \mathrm{NaOH}\) f. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) with \(0.120 \mathrm{M} \mathrm{NaOH}\)

A 0.1036 -g sample that contains only \(\mathrm{BaCl}_{2}\) and \(\mathrm{NaCl}\) is dissolved in \(50 \mathrm{~mL}\) of distilled water. Titrating with \(0.07916 \mathrm{M} \mathrm{AgNO}_{3}\) requires \(19.46 \mathrm{~mL}\) to reach the Fajans end point. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{BaCl}_{2}\) in the sample.

Prada and colleagues described an indirect method for determining sulfate in natural samples, such as seawater and industrial effluents. \({ }^{12}\) The method consists of three steps: precipitating the sulfate as \(\mathrm{PbSO}_{4}\); dissolving the \(\mathrm{PbSO}_{4}\) in an ammonical solution of excess EDTA to form the soluble \(\mathrm{Pb} \mathrm{Y}^{2-}\) complex; and titrating the excess EDTA with a standard solution of \(\mathrm{Mg}^{2+}\). The following reactions and equilibrium constants are known $$ \begin{array}{ll} \mathrm{PbSO}_{4}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) & K_{\mathrm{sp}}=1.6 \times 10^{-8} \\ \mathrm{~Pb}^{2+}(a q)+\mathrm{Y}^{4-}(a q) \rightleftharpoons \mathrm{Pb} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=1.1 \times 10^{18} \\ \mathrm{Mg}^{2+}(a q)+\mathrm{Y}^{4-}(a q) \rightleftharpoons \mathrm{Mg} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=4.9 \times 10^{8} \\ \mathrm{Zn}^{2+}(a q)+\mathrm{Y}^{4-}(a q)=\mathrm{Zn} \mathrm{Y}^{2-}(a q) & K_{\mathrm{f}}=3.2 \times 10^{16} \end{array} $$ (a) Verify that a precipitate of \(\mathrm{PbSO}_{4}\) will dissolve in a solution of \(\mathrm{Y}^{4-}\). (b) Sporek proposed a similar method using \(\mathrm{Zn}^{2+}\) as a titrant and found that the accuracy frequently was poor. \(^{13}\) One explanation is that \(\mathrm{Zn}^{2+}\) might react with the \(\mathrm{PbY}^{2-}\) complex, forming \(\mathrm{ZnY}^{2-}\). Show that this might be a problem when using \(\mathrm{Zn}^{2+}\) as a titrant, but that it is not a problem when using \(\mathrm{Mg}^{2+}\) as a titrant. Would such a displacement of \(\mathrm{Pb}^{2+}\) by \(\mathrm{Zn}^{2+}\) lead to the reporting of too much or too little sulfate? (c) In a typical analysis, a 25.00 -mL sample of an industrial effluent is carried through the procedure using \(50.00 \mathrm{~mL}\) of \(0.05000 \mathrm{M}\) EDTA. Titrating the excess EDTA requires \(12.42 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) \(\mathrm{Mg}^{2+}\). Report the molar concentration of \(\mathrm{SO}_{4}^{2-}\) in the sample of effluent.

Animal fats and vegetable oils are triesters formed from the reaction between glycerol \((1,2,3\) -propanetriol \()\) and three long-chain fatty acids. One of the methods used to characterize a fat or an oil is a determination of its saponification number. When treated with boiling aqueous \(\mathrm{KOH},\) an ester saponifies into the parent alcohol and fatty acids (as carboxylate ions). The saponification number is the number of milligrams of KOH required to saponify 1.000 gram of the fat or the oil. In a typical analysis a \(2.085-\mathrm{g}\) sample of butter is added to \(25.00 \mathrm{~mL}\) of \(0.5131 \mathrm{M} \mathrm{KOH}\). After saponification is complete the excess \(\mathrm{KOH}\) is back titrated with \(10.26 \mathrm{~mL}\) of \(0.5000 \mathrm{M} \mathrm{HCl}\). What is the saponification number for this sample of butter?

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