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Chapter 5: Problem 11

Express the following data for cations in solution as molarities. (a) \(14.2 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{L} ;\) (b) \(32.8 \mathrm{mg} \mathrm{K}^{+} / 100 \mathrm{mL};\) (c) \(225 \mu \mathrm{g} \mathrm{Zn}^{2+} / \mathrm{mL}\).

Short Answer

Expert verified
(a) 0.353 mM; (b) 8.38 M; and (c) 0.00344 M without clarification on total volume.

Step by step solution

01

Convert the mass into grams

First, it's needed to ensure that all quantities are given in the same unit, which is grams. This is easy for (a) because it's already given as milligrams, so one needs to divide it by 1000 to convert it into grams. For (b), the same operation should be done. For (c), the quantity is given in micrograms, so it should be divided by 1000000.
02

Convert the mass into moles

In the second step, the given mass should be converted into moles by dividing it by the molar mass of the element. The molar masses of \( \mathrm{Ca}^{2+} \), \( \mathrm{K}^{+}\) and \( \mathrm{Zn}^{2+}\) are 40.08 g/mol, 39.10 g/mol and 65.38 g/mol respectively.
03

Adjust the volume to liters

The final step is to adjust the volume for all given quantities into liters. For (a), the volume is given in liters hence no adjustments are needed. For (b), the volume is in ml and hence should be divided by 1000 to convert into liters. (c) has the volume given per ml; it is important to clarify the total volume of solution or continue with a calculation per milliliter.
04

Calculate the molarities

Finally, the molarity for each cation can be calculated by dividing the number of moles by the volume in liters.

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Key Concepts

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

Cation Concentration
Understanding cation concentration is crucial in solution chemistry. It refers to the number of moles of a positively charged ion in a given volume of solution. When you need to express concentration in molarity, you're essentially looking at how many moles of the cation are present in one liter of solution. To do this, follow these steps:

  • Convert the mass of the cation into grams, if necessary. This ensures uniformity in units.
  • Use the molar mass of the cation to convert the mass into moles.
  • Adjust the volume of the solution to liters to match the molarity definition.
  • Finally, calculate the molarity by dividing the number of moles of the cation by the volume in liters.
These steps help you find out how concentrated your solution is with a particular cation.
Molar Mass Conversion
Molar mass conversion is a key step in determining molarity. Every element has a specific molar mass, which is the mass of one mole of its atoms and is measured in g/mol. Here's how you convert mass to moles:

  • First, ensure the mass of the substance is in grams.
  • Next, use the molar mass of the element, such as Calcium (40.08 g/mol), Potassium (39.10 g/mol), or Zinc (65.38 g/mol).
  • Divide the mass of the cation by its molar mass to find the number of moles. For instance, if you have a mass of calcium, you take your grams of calcium and divide it by 40.08 g/mol.
This conversion allows you to determine how much of the cation you have in moles, paving the way for further calculations like molarity.
Solution Chemistry
Solution chemistry focuses on how substances interact within a solution, especially how they dissolve to form a homogeneous mixture. One core aspect is calculating molarity, which represents concentration. Solution chemistry requires paying attention to units and understanding volumes:

  • Always adjust volumes to liters when calculating molarities; this standardizes your calculations.
  • When given volumes in milliliters, convert them by dividing by 1000.
  • For tiny volume measurements like microliters, consider the total solution's volume or perform calculations per unit volume.
  • Molarity is then calculated using the formula \( M = \frac{moles}{liters} \).
Understanding the principles of solution chemistry helps in many real-world applications, from laboratories to industrial processes, ensuring precise formulation and reaction outcomes.

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Most popular questions from this chapter

When treated with dilute \(\mathrm{HCl}(\mathrm{aq}),\) the solid that reacts to produce a gas is (a) \(\mathrm{BaSO}_{3} ;\) (b) \(\mathrm{ZnO};\) (c) \(\mathrm{NaBr} ;\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\).

A sample of battery acid is to be analyzed for its sulfuric acid content. A \(1.00 \mathrm{mL}\) sample weighs \(1.239 \mathrm{g}\). This \(1.00 \mathrm{mL}\) sample is diluted to \(250.0 \mathrm{mL}\), and \(10.00 \mathrm{mL}\) of this diluted acid requires \(32.44 \mathrm{mL}\) of \(0.00498 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. What is the mass percent of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in the battery acid? (Assume that complete ionization and neutralization of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) occurs.)

For use in titrations, we want to prepare \(20 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about \(0.10 \mathrm{M}\) HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated \(\mathrm{HCl}(\mathrm{aq})\) \((d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })\) must be diluted with water to 20.0 L to prepare \(0.10 \mathrm{M} \mathrm{HCl}\) ? (b) \(\mathrm{A} 25.00\) \(\mathrm{mL}\) sample of the approximately \(0.10\) \(\mathrm{M}\) HCl prepared in part (a) requires \(20.93\) \(\mathrm{mL}\) of \(0.1186\) \(\mathrm{M}\) NaOH for its titration. What is the molarity of the \(\mathrm{HCl}(\mathrm{aq}) ?\) (c) Why is a titration necessary? That is, why not prepare a standard solution of \(0.1000\) \(\mathrm{M} \mathrm{HCl}\) simply by an appropriate dilution of the concentrated HCl(aq)?

A neutralization reaction between an acid and a base is a common method of preparing useful salts. Give net ionic equations showing how the following salts could be prepared in this way: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4};\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3} ;\) and \((\mathrm{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Assuming the volumes are additive, what is the \(\left[\mathrm{NO}_{3}^{-}\right]\) in a solution obtained by mixing \(275 \mathrm{mL}\) of \(0.283 \mathrm{M} \mathrm{KNO}_{3}, 328 \mathrm{mL}\) of \(0.421 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2},\) and \(784 \mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O} ?\)
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