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Chapter 4: Problem 122

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing \(10.0,25.0,50.0,75.0\), and \(100 . \mathrm{ppm}\) of copper from a commercially produced \(1000.0\) -ppm solution? Assume each solution has a final volume of \(100.0 \mathrm{~mL}\). (See Exercise 121 for definitions.)

Short Answer

Expert verified
To prepare standard solutions of copper with concentrations 10.0, 25.0, 50.0, 75.0, and 100.0 ppm from a 1000 ppm solution, use the dilution formula \(C_1 V_1 = C_2 V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume. Calculate the required volume of the 1000 ppm solution for each desired concentration (\(V_1\)), then measure this amount using a calibrated pipette, transfer it into a separate volumetric flask, and dilute to the 100.0 mL mark with an appropriate solvent (usually water). Make sure to mix well to achieve a uniformly distributed copper solution in the desired concentration.

Step by step solution

01

Calculate the Volume of the Initial Solution

Start by plugging the given concentration and the desired concentration of copper into the formula mentioned above. Solve for the volume of the initial solution, \(V_1\): \( V_1 = \frac{C_2 V_2}{C_1} \) Note that: - \(C_1\) = 1000 ppm (given) - \(C_2\) = Desired concentration (10.0, 25.0, 50.0, 75.0, and 100.0 ppm) - \(V_2\) = Final volume (given as 100.0 mL)
02

Measure the Calculated Volume of the Initial Solution

Once the volume of the 1000 ppm initial solution is calculated for each desired concentration, use a calibrated pipette to precisely measure and transfer that volume into a separate container, often a volumetric flask.
03

Dilute to the Final Volume with Solvent

Add the appropriate solvent (usually water) to the container up to the 100.0 mL mark. Make sure to mix well to ensure that the desired concentration of copper is evenly distributed in the solution. Now we can put this process into practice for each desired concentration: a) Desired concentration: \(10.0\) ppm
04

Calculate the Volume of the Initial Solution

\( V_1 = \frac{(10.0 \,\text{ppm})(100.0\, \text{mL})}{1000.0\, \mathrm{ppm}} = 1.0\, \mathrm{mL} \) b) Desired concentration: \(25.0\) ppm
05

Calculate the Volume of the Initial Solution

\( V_1 = \frac{(25.0\, \text{ppm})(100.0\, \text{mL})}{1000.0\, \mathrm{ppm}} = 2.5\, \mathrm{mL} \) c) Desired concentration: \(50.0\) ppm
06

Calculate the Volume of the Initial Solution

\( V_1 = \frac{(50.0\, \text{ppm})(100.0\, \text{mL})}{1000.0\, \mathrm{ppm}} = 5.0\, \mathrm{mL} \) d) Desired concentration: \(75.0\) ppm
07

Calculate the Volume of the Initial Solution

\( V_1 = \frac{(75.0\, \text{ppm})(100.0\, \text{mL})}{1000.0\, \mathrm{ppm}} = 7.5\, \mathrm{mL} \) e) Desired concentration: \(100.0\) ppm
08

Calculate the Volume of the Initial Solution

\( V_1 = \frac{(100.0\, \text{ppm})(100.0\, \text{mL})}{1000.0\, \mathrm{ppm}} = 10.0\, \mathrm{mL} \) For each of these solutions, follow Steps 2 and 3 to prepare the desired concentration of copper standard solutions.

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Most popular questions from this chapter

A \(10.00-\mathrm{mL}\) sample of vinegar, an aqueous solution of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\), is titrated with \(0.5062 \mathrm{M} \mathrm{NaOH}\), and \(16.58 \mathrm{~mL}\) is required to reach the equivalence point. a. What is the molarity of the acetic acid? b. If the density of the vinegar is \(1.006 \mathrm{~g} / \mathrm{cm}^{3}\), what is the mass percent of acetic acid in the vinegar?

A \(0.500-\mathrm{L}\) sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution was analyzed by taking a 100.0-mL aliquot and adding \(50.0 \mathrm{~mL}\) of \(0.213 \mathrm{M} \mathrm{NaOH}\). After the reaction occurred, an excess of \(\mathrm{OH}^{-}\) ions remained in the solution. The excess base required \(13.21 \mathrm{~mL}\) of \(0.103 M\) \(\mathrm{HCl}\) for neutralization. Calculate the molarity of the original sample of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Sulfuric acid has two acidic hydrogens.

Using the general solubility rules given in Table 4.1, name three reagents that would form precipitates with each of the following ions in aqueous solution. Write the net ionic equation for each of your suggestions. a. chloride ion b. calcium ion c. iron(III) ion d. sulfate ion e. mercury(I) ion, \(\mathrm{Hg}_{2}{ }^{2+}\) f. silver ion

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains two acidic hydrogens

A \(2.20-\mathrm{g}\) sample of an unknown acid (empirical formula = \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\) ) is dissolved in \(1.0 \mathrm{~L}\) of water. A titration required \(25.0 \mathrm{~mL}\) of \(0.500 M \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?
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