Chapter 4: Problem 19
What is an acid and what is a base? An acid-base reaction is sometimes called a proton-transfer reaction. Explain.
Chapter 4: Problem 19
What is an acid and what is a base? An acid-base reaction is sometimes called a proton-transfer reaction. Explain.
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Get started for freeCalculate the molarity of each of these solutions. a. A \(5.623-\mathrm{g}\) sample of \(\mathrm{NaHCO}_{3}\) is dissolved in enough water to make \(250.0 \mathrm{~mL}\) of solution. b. A \(184.6-\mathrm{mg}\) sample of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is dissolved in enough water to make \(500.0 \mathrm{~mL}\) of solution. c. A 0.1025-g sample of copper metal is dissolved in \(35 \mathrm{~mL}\) of concentrated \(\mathrm{HNO}_{3}\) to form \(\mathrm{Cu}^{2+}\) ions and then water is added to make a total volume of \(200.0 \mathrm{~mL}\). (Calculate the molarity of \(\mathrm{Cu}^{2+} .\) )
A solution was prepared by mixing \(50.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) and \(100.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the molarity of the final solution of nitric acid.
Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 \mathrm{M} \mathrm{NaOH}\) from \(1.00 \mathrm{M} \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr} \mathrm{O}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution
The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{~mL}\), which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10\) ppm DDT \(\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)
The drawings below represent aqueous solutions. Solution A is \(2.00 \mathrm{~L}\) of a \(2.00 \mathrm{M}\) aqueous solution of copper(II) nitrate. Solution \(\mathrm{B}\) is \(2.00 \mathrm{~L}\) of a \(3.00 \mathrm{M}\) aqueous solution of potassium hydroxide. a. Draw a picture of the solution made by mixing solutions \(\mathrm{A}\) and \(\mathrm{B}\) together after the precipitation reaction takes place. Make sure this picture shows the correct relative volume compared to solutions \(\mathrm{A}\) and \(\mathrm{B}\), and the correct relative number of ions, along with the correct relative amount of solid formed. b. Determine the concentrations (in \(M\) ) of all ions left in solution (from part a) and the mass of solid formed.
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