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Chapter 11: Problem 33

A \(50.0-\mathrm{mL}\) aliquot of solution containing \(0.450 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\) (FM 120.37) in \(0.500 \mathrm{~L}\) required \(37.6 \mathrm{~mL}\) of EDTA solution for titration. How many milligrams of \(\mathrm{CaCO}_{3}\) (FM \(100.09\) ) will react with \(1.00 \mathrm{~mL}\) of this EDTA solution?

Short Answer

Expert verified
0.998 mg of \(\mathrm{CaCO}_{3}\) will react with 1.00 mL of the EDTA solution.

Step by step solution

01

Calculate Moles of MgSO4

First, calculate the moles of \(\mathrm{MgSO}_{4}\) in the 50.0 mL aliquot. We know that there is 0.450 g in 500.0 mL, so we must scale this down to 50.0 mL. Use the formula: \[ \text{moles of } \mathrm{MgSO}_{4} = \frac{0.450 \text{ g}}{120.37 \text{ g/mol}} \times \frac{50.0 \text{ mL}}{500.0 \text{ mL}} \].Upon calculation, the moles of \(\mathrm{MgSO}_{4}\) are approximately 0.000375 moles.
02

Determine Moles of EDTA

Since the 50.0 mL aliquot required 37.6 mL of EDTA for titration, and assuming a 1:1 molar ratio between \(\mathrm{MgSO}_{4}\) and EDTA, the moles of EDTA required are equal to the moles of \(\mathrm{MgSO}_{4}\).Thus, 0.000375 moles of EDTA are present in 37.6 mL of solution.
03

Calculate Molarity of EDTA Solution

Now, calculate the molarity of the EDTA solution using the moles determined:\[ M = \frac{\text{moles of EDTA}}{\text{volume of EDTA in L}} = \frac{0.000375}{0.0376} \approx 0.00998 \text{ M} \].So, the molarity of the EDTA solution is approximately 0.00998 M.
04

Calculate Moles of EDTA in 1 mL

Using the molarity of the EDTA solution, calculate the moles of EDTA in 1.00 mL of solution:\[ \text{moles of EDTA in 1 mL} = 0.00998 \text{ M} \times 0.001 \text{ L} = 0.00000998 \text{ moles} \].
05

Calculate Mass of CaCO3 Reacting with 1 mL of EDTA Solution

Assuming a 1:1 molar reaction between EDTA and \(\mathrm{CaCO}_{3}\), the moles of \(\mathrm{CaCO}_{3}\) will be the same as the moles of EDTA. Calculate the mass:\[ \text{mass of } \mathrm{CaCO}_{3} = 0.00000998 \text{ mol} \times 100.09 \text{ g/mol} = 0.000998 \text{ g} = 0.998 \text{ mg} \].

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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 common laboratory technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration, called the titrant. In this method, the titrant is slowly added to the unknown solution until the reaction reaches its endpoint. This endpoint is often identified using a color change from an indicator or a pH meter reading.
Key points to remember about titration:
  • It relies on a complete reaction between reactants.
  • Precise measurement of volumes is crucial for accurate results.
  • Equipment like burettes, pipettes, and flasks are used for carrying out titrations.
In the original exercise, titration was employed to determine how much calcium carbonate (\(\mathrm{CaCO}_{3}\)) would react with 1 mL of EDTA solution. This specific type of titration helps in quantitative chemical analysis by ensuring accurate calculations of molar concentration.
Molarity Calculation Explained
Molarity, often denoted by "M," is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is:\[M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\]To illustrate, in the exercise, the molarity of the EDTA solution was determined using the calculated moles of EDTA in a specific volume of solution. For the calculated moles of \(0.000375\) and a volume of \(37.6\) mL converted to liters (\(0.0376\) L), the molarity came out to be approximately \(0.00998 \text{ M}\).
Important aspects of molarity calculation:
  • Ensure units are consistent (moles and liters).
  • Careful measurement is necessary for accuracy.
  • Understanding the relationship between molarity and molar mass can simplify conversions and calculations.
Chemical Stoichiometry and Its Importance
Chemical stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It ensures that the equation governing a chemical reaction is balanced, meaning the number of atoms for each element is the same on both sides of the equation. In stoichiometry calculations:
  • Mole ratios from the balanced chemical equation are used.
  • Conversion between moles, mass, and volume often occurs.
  • Precise stoichiometric coefficients provide insight into reaction dynamics.
In the given problem, the 1:1 stoichiometric relationship between \(\mathrm{MgSO}_{4}\) and EDTA, as well as between EDTA and \(\mathrm{CaCO}_{3}\), was crucial. This implied that for every mole of one reactant, an equal mole of the other was necessary. Thus, knowing these ratios allowed for the calculation of the mass of \(\mathrm{CaCO}_{3}\) reacting with EDTA.
Understanding EDTA Titration
EDTA titration is a specific type of complexometric titration used particularly for measuring concentrations of metal ions in solution. EDTA (Ethylenediaminetetraacetic acid) is a chelating agent that binds specifically with metal ions to form stable complexes. This property makes it perfect for analyzing solutions with metal ions like calcium and magnesium. Characteristics of EDTA titration:
  • Used extensively for water hardness testing by detecting calcium and magnesium ions.
  • Relies on indicators such as Eriochrome Black T that change color during endpoint detection.
  • Can determine concentration through the use of stoichiometrically balanced equations.
The exercise demonstrates an EDTA titration where EDTA reacts with magnesium sulfate initially, then calculation proceeds to deduce the interaction with calcium carbonate, showcasing quantitative chemical analysis using this titration method.

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Most popular questions from this chapter

Give an example of the use of a masking agent.

A \(50.0-\mathrm{mL}\) sample containing \(\mathrm{Ni}^{2+}\) was treated with \(25.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M}\) EDTA to complex all the \(\mathrm{Ni}^{2+}\) and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring \(5.00 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{Zn}^{2+}\). What was the concentration of \(\mathrm{Ni}^{2+}\) in the original solution?

Microequilibrium constants for binding of metal to a protein. The iron- transport protein, transferrin, has two distinguishable metal-binding sites, designated a and b. The microequilibrium formation constants for each site are defined as follows: For example, the formation constant \(k_{\mathrm{la}}\) refers to the reaction \(\mathrm{Fe}^{3+}+\) transferrin \(\rightleftharpoons \mathrm{Fe}_{\mathrm{a}}\) transferrin, in which the metal ion binds to site a: $$ k_{1 \mathrm{a}}=\frac{\left[\mathrm{Fe}_{a} \text { transferrin }\right]}{\left[\mathrm{Fe}^{3+}\right][\text { transferrin }]} $$ (a) Write the chemical reactions corresponding to the conventional macroscopic formation constants, \(K_{1}\) and \(K_{2}\). (b) Show that \(K_{1}=k_{1 \mathrm{a}}+k_{1 \mathrm{~b}}\) and \(K_{2}^{-1}=k_{2 \mathrm{a}}^{-1}+k_{2 \mathrm{~b}}^{-1}\). (c) Show that \(k_{1 \mathrm{a}} k_{2 \mathrm{~b}}=k_{1 \mathrm{~b}} k_{2 \mathrm{a}}\). This expression means that, if you know any three of the microequilibrium constants, you automatically know the fourth one. (d) A challenge to your sanity. From the equilibrium constants below, find the equilibrium fraction of each of the four species in the diagram above in circulating blood that is \(40 \%\) saturated with iron (that is, Fe/transferrin \(=0.80\), because each protein can bind 2 Fe). Effective formation constants for blood plasma at \(\mathrm{pH} 7.4\) \begin{tabular}{ll} \hline\(k_{1 \mathrm{a}}=6.0 \times 10^{22}\) & \(k_{2 \mathrm{a}}=2.4 \times 10^{22}\) \\ \(k_{1 \mathrm{~b}}=1.0 \times 10^{22}\) & \(k_{2 \mathrm{~b}}=4.2 \times 10^{21}\) \\ \(K_{1}=7.0 \times 10^{22}\) & \(K_{2}=3.6 \times 10^{21}\) \\ \hline \end{tabular} The binding constants are so large that you may assume that there is negligible free \(\mathrm{Fe}^{3+}\). To get started, let's use the abbreviations [T] = \([\) transferrin \(],[\mathrm{FeT}]=\left[\mathrm{Fe}_{\mathrm{a}} \mathrm{T}\right]+\left[\mathrm{Fe}_{\mathrm{b}} \mathrm{T}\right]\), and \(\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=\left[\mathrm{Fe}_{2} \operatorname{transferrin}\right] .\) Now we can write Mass balance for protein: \(\quad[\mathrm{T}]+[\mathrm{FeT}]+\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=1\) Mass balance for iron: $$ \frac{[\mathrm{FeT}]+2\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}{[\mathrm{T}]+[\mathrm{Fe} \mathrm{T}]+\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}=[\mathrm{FeT}]+2\left[\mathrm{Fe}_{2} \mathrm{~T}\right]=0.8 $$ Combined equilibria: \(\frac{K_{1}}{K_{2}}=19_{.44}=\frac{[\mathrm{FeT}]^{2}}{[\mathrm{~T}]\left[\mathrm{Fe}_{2} \mathrm{~T}\right]}\) Now you have three equations with three unknowns and should be able to tackle this problem.

Calculate \(\mathrm{pCo}^{2+1}\) at each of the following points in the titration of \(25.00 \mathrm{~mL}\) of \(0.02026 \mathrm{M} \mathrm{Co}^{2+}\) by \(0.03855 \mathrm{M}\) EDTA at \(\mathrm{pH}\) 6.00: (a) \(12.00 \mathrm{~mL}\); (b) \(V_{c}\); (c) \(14.00 \mathrm{~mL}\). 11-8. Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M} \mathrm{MnSO}_{4}\) with \(0.0100 \mathrm{M}\) EDTA in a solution buffered to \(\mathrm{pH} 8.00\). Calculate \(\mathrm{pMn}^{2+}\) at the following volumes of added EDTA and sketch the titration curve: (a) \(0 \mathrm{~mL}\) (d) \(49.0 \mathrm{~mL}\) (g) \(50.1 \mathrm{~mL}\) (b) \(20.0 \mathrm{~mL}\) (e) \(49.9 \mathrm{~mL}\) (h) \(55.0 \mathrm{~mL}\) (c) \(40.0 \mathrm{~mL}\) (f) \(50.0 \mathrm{~mL}\) (i) \(60.0 \mathrm{~mL}\)

A \(1.000-\mathrm{mL}\) sample of unknown containing \(\mathrm{Co}^{2+}\) and \(\mathrm{Ni}^{2+}\) was treated with \(25.00 \mathrm{~mL}\) of \(0.03872 \mathrm{M}\) EDTA. Back titration with \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) at \(\mathrm{pH} 5\) required \(23.54 \mathrm{~mL}\) to reach the xylenol orange end point. A \(2.000-\mathrm{mL}\) sample of unknown was passed through an ion- exchange column that retards \(\mathrm{Co}^{2+}\) more than \(\mathrm{Ni}^{2+}\). The \(\mathrm{Ni}^{2+}\) that passed through the column was treated with \(25.00 \mathrm{~mL}\) of \(0.03872 \mathrm{M}\) EDTA and required \(25.63 \mathrm{~mL}\) of \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) for back titration. The \(\mathrm{Co}^{2+}\) emerged from the column later. It, too, was treated with \(25.00 \mathrm{~mL}\) of \(0.03872\) M EDTA. How many milliliters of \(0.02127 \mathrm{M} \mathrm{Zn}^{2+}\) will be required for back titration?
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