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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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Most popular questions from this chapter

Write a balanced equation for the redox reactions. (a) The reaction of aluminum metal with hydroiodic acid. (b) The reduction of vanadyl ion ( \(\mathrm{VO}^{2+}\) ) to vanadic ion \(\left(\mathrm{V}^{3+}\right)\) in acidic solution with zinc metal as the reducing agent. (c) The oxidation of methanol by chlorate ion in acidic solution, producing carbon dioxide gas, water, and chlorine dioxide gas as products.

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

A \(25.00 \mathrm{mL}\) sample of \(0.132 \mathrm{M}\) \(\mathrm{HNO}_{3}\) is mixed with \(10.00 \mathrm{mL}\) of \(0.318 \mathrm{M} \mathrm{KOH} .\) Is the resulting solution acidic, basic, or exactly neutralized?

Determine which of the following react(s) with HCl(ag) to produce a gas, and write a net ionic equation(s) for the reaction(s). (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4} ;\) (b) \(\mathrm{KHSO}_{3}\); (c) \(\mathrm{Zn}(\mathrm{OH})_{2};\) (d) \(\mathrm{CaCl}_{2}\).
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