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Chapter 9: Problem 37

After removing the membranes from an eggshell, the shell is dried and its mass recorded as \(5.613 \mathrm{~g} .\) The eggshell is transferred to a \(250-\mathrm{mL}\) beaker and dissolved in \(25 \mathrm{~mL}\) of \(6 \mathrm{M}\) HCl. After filtering, the solution that contains the dissolved eggshell is diluted to \(250 \mathrm{~mL}\) in a volumetric flask. A \(10.00-\mathrm{mL}\) aliquot is placed in a \(125-\mathrm{mL}\) Erlenmeyer flask and buffered to a \(\mathrm{pH}\) of 10 . Titrating with \(0.04988 \mathrm{M}\) EDTA requires \(44.11 \mathrm{~mL}\) to reach the end point. Determine the amount of calcium in the eggshell as \(\% \mathrm{w} / \mathrm{w} \mathrm{CaCO}_{3}\).

Short Answer

Expert verified
The calcium content is 98.03% w/w as CaCO3.

Step by step solution

01

Calculate moles of EDTA used

To find the moles of EDTA used, multiply its concentration by the volume used (in liters):\[ \text{moles of EDTA} = 0.04988 \text{ M} \times 44.11 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.0021993 \text{ moles} \]
02

Relate moles of EDTA to moles of Calcium

In a titration involving EDTA and calcium, the ratio is 1:1, meaning 1 mole of EDTA reacts with 1 mole of calcium ions. Thus, moles of calcium ions = moles of EDTA = 0.0021993 moles.
03

Calculate moles of Calcium in the entire solution

The calcium calculation in the aliquot can be scaled to the entire solution:\[ \text{moles of } Ca = 0.0021993 \text{ moles} \times \frac{250 \text{ mL}}{10 \text{ mL}} = 0.0549825 \text{ moles} \]
04

Convert moles of Calcium to mass of CaCO3

Use the molar mass of \( \mathrm{CaCO}_3 \), which is \(100.09 \text{ g/mol}\), to convert the moles of calcium to grams of \(\mathrm{CaCO}_3\):\[ \text{mass of } \mathrm{CaCO}_3 = 0.0549825 \text{ moles} \times 100.09 \text{ g/mol} = 5.502 \text{ grams} \]
05

Calculate percentage of CaCO3 in eggshell

The percentage of \(\mathrm{CaCO}_3\) is given by the ratio of the mass of \(\mathrm{CaCO}_3\) to the total mass of the eggshell, multiplied by 100:\[ \text{Percentage of } \mathrm{CaCO}_3 = \left(\frac{5.502 \text{ g}}{5.613 \text{ g}}\right) \times 100 = 98.03\% \]

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Key Concepts

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

Understanding Titration
Titration is a fundamental technique in analytical chemistry used to determine the unknown concentration of a substance. It involves gradually adding a solution of known concentration, called the titrant, to a solution of unknown concentration until a reaction is complete. When performing a titration, key terms to understand are the "end point" and "equivalence point".
  • End Point: The moment when the titration is stopped, often indicated by a color change if an indicator is used.
  • Equivalence Point: When the amount of titrant is chemically equivalent to the substance being titrated.
To ensure accurate results, it is important that the end point closely matches the equivalence point. In our exercise, the titration of calcium ions requires the use of EDTA as the titrant. The manner in which EDTA binds metals, including calcium, is key to this process.
Determining Calcium in Eggshells
Determining the calcium content in an eggshell is a practical application of analytical chemistry. Eggs are primarily made up of calcium carbonate \( \text{CaCO}_3 \), which, in this context, provides a measurable source of calcium. To find the calcium percentage, the eggshell undergoes several processing steps:
  • The eggshell is dissolved in hydrochloric acid \( \text{HCl} \), releasing calcium ions into solution.
  • The solution is filtered and diluted, ensuring even distribution of calcium ions.
  • A portion, or aliquot, of this solution is buffered to a \( \text{pH} \) of 10 to prepare for titration.
The method leverages the relationship between calcium and EDTA, which, in a neutral to basic solution, forms a stable complex. Thus, the amount of EDTA used reflects the exact quantity of calcium in the solution. Scaling the reaction to the entire sample yields the total calcium content, allowing for calculation of calcium carbonate percentage.
Role of EDTA in Complexometric Titration
EDTA, or ethylenediaminetetraacetic acid, is a versatile chelating agent used in titrations to quantify metal ions. In complexometric titration, EDTA acts by binding to metal ions, forming stable, water-soluble complexes. The key benefit of using EDTA is its ability to form a 1:1 ratio with most metal ions, including calcium.
  • Complex Formation: When EDTA is added to a solution containing calcium ions, it forms a complex \( \text{Ca-EDTA} \). This process consumes a stoichiometric amount of EDTA, reflecting the calcium amount.
  • Indicator: Often, an indicator such as Eriochrome Black T is used to signal the end point by changing color once all metal ions are bound.
In the titration scenario described in the exercise, knowing the precise concentration of EDTA allows the calculation of calcium ion concentration. Because the EDTA to calcium ratio is 1:1, the moles of EDTA used directly translate to the moles of calcium present. This makes EDTA an invaluable tool in quantitative analysis of metal ions.

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Most popular questions from this chapter

The amount of uranium in an ore is determined by an indirect redox titration. The analysis is accomplished by dissolving the ore in sulfuric acid and reducing \(\mathrm{UO}_{2}^{2+}\) to \(\mathrm{U}^{4+}\) with a Walden reductor. The solution is treated with an excess of \(\mathrm{Fe}^{3+}\), forming \(\mathrm{Fe}^{2+}\) and \(\mathrm{U}^{6+}\). The \(\mathrm{Fe}^{2+}\) is titrated with a standard solution of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} .\) In a typical analysis a 0.315 -g sample of ore is passed through the Walden reductor and treated with \(50.00 \mathrm{~mL}\) of \(0.0125 \mathrm{M} \mathrm{Fe}^{3+}\). Back titrating with 0.00987 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) requires \(10.52 \mathrm{~mL}\). What is the \(\% \mathrm{w} / \mathrm{w} \mathrm{U}\) in the sample?

Under basic conditions, \(\mathrm{MnO}_{4}^{-}\) is used as a titrant for the analysis of \(\mathrm{Mn}^{2+}\), with both the analyte and the titrant forming \(\mathrm{MnO}_{2}\). In the analysis of a mineral sample for manganese, a \(0.5165-\mathrm{g}\) sample is dissolved and the manganese reduced to \(\mathrm{Mn}^{2+}\). The solution is made basic and titrated with \(0.03358 \mathrm{M} \mathrm{KMnO}_{4}\), requiring \(34.88 \mathrm{~mL}\) to reach the end point. Calculate the \(\% \mathrm{w} / \mathrm{w} \mathrm{Mn}\) in the mineral sample.

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 \(\mathrm{NaCN}\).

Calculate or sketch the titration curve for a \(50.0 \mathrm{~mL}\) solution of a \(0.100 \mathrm{M}\) monoprotic weak acid \(\left(\mathrm{p} K_{\mathrm{a}}=8.0\right)\) with \(0.1 \mathrm{M}\) strong base in a nonaqueous solvent with \(K_{\mathrm{s}}=10^{-20}\). You may assume that the change in solvent does not affect the weak acid's \(\mathrm{p} K_{\mathrm{a}}\). Compare your titration curve to the titration curve when water is the solvent.

A variety of systematic and random errors are possible when standardizing a solution of \(\mathrm{NaOH}\) against the primary weak acid standard potassium hydrogen phthalate (KHP). Identify, with justification, whether the following are sources of systematic error or random error, or if they have no affect on the error. If the error is systematic, then indicate whether the experimentally determined molarity for \(\mathrm{NaOH}\) is too high or too low. The standardization reaction is $$ \mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) The balance used to weigh KHP is not properly calibrated and always reads \(0.15 \mathrm{~g}\) too low. (b) The indicator for the titration changes color between a \(\mathrm{pH}\) of \(3-4\). (c) An air bubble, which is lodged in the buret's tip at the beginning of the analysis, dislodges during the titration. (d) Samples of KHP are weighed into separate Erlenmeyer flasks, but the balance is tarred only for the first flask. (e) The KHP is not dried before it is used. (f) The \(\mathrm{NaOH}\) is not dried before it is used. (g) The procedure states that the sample of KHP should be dissolved in \(25 \mathrm{~mL}\) of water, but it is accidentally dissolved in \(35 \mathrm{~mL}\) of water.
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