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How many milliliters of \(0.0500 \mathrm{M}\) EDTA are required to react with \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{Ca}^{2+} ?\) With \(50.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{Al}^{3+} ?\)

Short Answer

Expert verified
10 mL of 0.0500 M EDTA is needed for both Ca2+ and Al3+ solutions.

Step by step solution

01

Determine the moles of Ca2+ and Al3+

Calculate the number of moles of each ion present in the solutions. Since the concentration of both solutions is given as 0.0100 M and the volume is 50.0 mL (or 0.0500 L), use the formula:\[ n = C \times V \]For Ca\(^{2+}\):\[ n_{Ca^{2+}} = 0.0100 \, M \times 0.0500 \, L = 0.000500 \, ext{mol} \]For Al\(^{3+}\):\[ n_{Al^{3+}} = 0.0100 \, M \times 0.0500 \, L = 0.000500 \, ext{mol} \]
02

Write the balanced reactions with EDTA

EDTA forms 1:1 complexes with metal ions. The balanced equations are:1. For Ca\(^{2+}\):\[ ext{Ca}^{2+} + ext{EDTA}^{4-} \rightarrow [ ext{CaEDTA}]^{2-} \]2. For Al\(^{3+}\):\[ ext{Al}^{3+} + ext{EDTA}^{4-} \rightarrow [ ext{AlEDTA}]^{3-} \]
03

Calculate the volume of EDTA solution needed

Since the reaction is 1:1, the number of moles of EDTA required will be equal to the number of moles of the metal ion. Use the concentration of EDTA to find the volume:For Ca\(^{2+}\):\[ V_{EDTA} = \frac{n_{Ca^{2+}}}{C_{EDTA}} = \frac{0.000500 \, ext{mol}}{0.0500 \, M} = 0.0100 \, ext{L} = 10.0 \, ext{mL} \]For Al\(^{3+}\):\[ V_{EDTA} = \frac{n_{Al^{3+}}}{C_{EDTA}} = \frac{0.000500 \, ext{mol}}{0.0500 \, M} = 0.0100 \, ext{L} = 10.0 \, ext{mL} \]
04

Conclusion

For both the Ca\(^{2+}\) and Al\(^{3+}\) solutions, the same volume of 0.0500 M EDTA is required (10.0 mL) because the moles of metal ions are equal and both react with EDTA on a 1:1 basis.

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Key Concepts

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

Metal Ion Concentration
Understanding metal ion concentration is key to calculating how much EDTA is needed for a titration. Concentration, given in molarity (M), tells us how many moles of a substance are present in one liter of solution.

In our problem, both calcium (\(\text{Ca}^{2+}\)) and aluminum (\(\text{Al}^{3+}\)) ions have a concentration of \(0.0100 \, \mathrm{M}\). This indicates that there are \(0.0100\) moles of each ion in \(1\) liter of their respective solutions. Using the relationship between moles, concentration, and volume, we can determine the amount of metal ions present in any volume of solution.

For the given \(50.0 \, \text{mL}\) (or \(0.0500 \, \text{L}\)) of each solution, multiply the molarity by the volume to find the number of moles of ions:
  • For calcium: \(n_{\text{Ca}^{2+}} = 0.0100 \, \text{M} \times 0.0500 \, \text{L} = 0.000500 \, \text{mol}\)
  • For aluminum: \(n_{\text{Al}^{3+}} = 0.0100 \, \text{M} \times 0.0500 \, \text{L} = 0.000500 \, \text{mol}\)
These calculations demonstrate that both solutions contain the same number of moles of metal ions.
1:1 Complex Formation
EDTA is a versatile chelating agent that forms strong complexes with metal ions in a 1:1 molar ratio, meaning one mole of EDTA reacts with one mole of metal ion. This is crucial in our calculations because it simplifies the process of determining how much EDTA is necessary.

For calcium ions (\(\text{Ca}^{2+}\)), the reaction with EDTA is:
  • \(\text{Ca}^{2+} + \text{EDTA}^{4-} \rightarrow [\text{CaEDTA}]^{2-}\)
Similarly, for aluminum ions (\(\text{Al}^{3+}\)), the reaction is:
  • \(\text{Al}^{3+} + \text{EDTA}^{4-} \rightarrow [\text{AlEDTA}]^{3-}\)
In both cases, the stoichiometry, or the mole ratio of reactants to products, is 1:1. This means that for every mole of \(\text{Ca}^{2+}\) or \(\text{Al}^{3+}\), one mole of EDTA is required to completely react, forming a stable complex. This concept is fundamental in EDTA titrations, as it allows easy calculation of the needed EDTA solution volume, directly based on the amount of metal ion present.
Molarity Calculations
Molarity is a measure of the concentration of a solute in a solution, defined as moles of solute per liter of solution. Calculating molarity helps in understanding reactions in terms of how much reactant is present to react or titrate.

Here's how we calculate the volume of EDTA solution required for our metal ions: we know the amount of metal ions (moles) and the concentration of EDTA (molarity) is given as \(0.0500 \, \mathrm{M}\).
  • For both \(\text{Ca}^{2+}\) and \(\text{Al}^{3+}\) ions: \(\text{Volume of EDTA required} = \frac{\text{moles of metal ion}}{\text{molarity of EDTA}}\)
  • Since the moles of metal ion are \(0.000500\), the calculation for volume of EDTA is:\[V_{\text{EDTA}} = \frac{0.000500 \, \text{mol}}{0.0500 \, \text{M}} = 0.0100 \, \text{L} = 10.0 \, \text{mL}\]
This simple division gives the volume in liters; converting to milliliters gives the practical volume needed for the reaction. This approach helps efficiently solve titration problems by combining knowledge of molarity with stoichiometry.
Chemical Reactions
In EDTA titration, understanding chemical reactions and their stoichiometry is vital. A chemical equation represents how substances interact and transform, showing us the proportions in which they react or are consumed.

For our titration problem, the chemical reactions with EDTA involve a single-step process where the calcium or aluminum ions bind to EDTA, forming a metal-EDTA complex.
  • For calcium: \(\text{Ca}^{2+} + \text{EDTA}^{4-} \rightarrow [\text{CaEDTA}]^{2-}\)
  • For aluminum: \(\text{Al}^{3+} + \text{EDTA}^{4-} \rightarrow [\text{AlEDTA}]^{3-}\)
The core idea here is that EDTA binds the metal ion in a 1:1 ratio, forming a stable and soluble chelate complex in solution. Understanding and balancing these chemical equations is crucial as it informs us exactly how much EDTA is required to titrate a given amount of metal ions.
These reactions highlight the use of EDTA as a chelating agent in complexometric titrations, wherein the balanced chemical equation guides the stoichiometric calculations necessary for quantitative analysis.

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Most popular questions from this chapter

The sulfur content of insoluble sulfides that do not readily dissolve in acid can be measured by oxidation with \(\mathrm{Br}_{2}\) to \(\mathrm{SO}_{4}^{2-} .25\) Metal ions are then replaced with \(\mathrm{H}^{+}\)by an ion-exchange column, and sulfate is precipitated as \(\mathrm{BaSO}_{4}\) with a known excess of \(\mathrm{BaCl}_{2}\). The excess \(\mathrm{Ba}^{2+}\) is then titrated with EDTA to determine how much was present. (To make the indicator end point clearer, a small, known quantity of \(\mathrm{Zn}^{2+}\) also is added. The EDTA titrates both the \(\mathrm{Ba}^{2+}\) and the \(\mathrm{Zn}^{2+}\).) Knowing the excess \(\mathrm{Ba}^{2+}\), we can calculate how much sulfur was in the original material. To analyze the mineral sphalerite ( \(\mathrm{ZnS}\), FM 97.46), \(5.89 \mathrm{mg}\) of powdered solid were suspended in a mixture of \(\mathrm{CCl}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) containing \(1.5 \mathrm{mmol} \mathrm{Br}_{2}\). After \(1 \mathrm{~h}\) at \(20^{\circ} \mathrm{C}\) and \(2 \mathrm{~h}\) at \(50^{\circ} \mathrm{C}\), the powder dissolved and the solvent and excess \(\mathrm{Br}_{2}\) were removed by heating. The residue was dissolved in \(3 \mathrm{~mL}\) of water and passed through an ion-exchange column to replace \(\mathrm{Zn}^{2+}\) with \(\mathrm{H}^{+}\). Then \(5.000 \mathrm{~mL}\) of \(0.01463 \mathrm{M} \mathrm{BaCl}_{2}\) were added to precipitate all sulfate as \(\mathrm{BaSO}_{4}\). After the addition of \(1.000 \mathrm{~mL}\) of \(0.01000 \mathrm{M} \mathrm{ZnCl}_{2}\) and \(3 \mathrm{~mL}\) of ammonia buffer, \(\mathrm{pH} 10\), the excess \(\mathrm{Ba}^{2+}\) and \(\mathrm{Zn}^{2+}\) required \(2.39 \mathrm{~mL}\) of \(0.00963 \mathrm{M}\) EDTA to reach the Calmagite end point. Find the weight percent of sulfur in the sphalerite. What is the theoretical value?

Describe what is done in a displacement titration and give an example.

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?

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?

A \(50.0-\mathrm{mL}\) solution containing \(\mathrm{Ni}^{2+}\) and \(\mathrm{Zn}^{2+}\) was treated with \(25.0 \mathrm{~mL}\) of \(0.0452 \mathrm{M}\) EDTA to bind all the metal. The excess unreacted EDTA required \(12.4 \mathrm{~mL}\) of \(0.0123 \mathrm{M} \mathrm{Mg}^{2+}\) for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another \(29.2 \mathrm{~mL}\) of \(\mathrm{Mg}^{2+}\) were required for reaction with the liberated EDTA. Calculate the molarity of \(\mathrm{Ni}^{2+}\) and \(\mathrm{Zn}^{2+}\) in the original solution.

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