Chapter 17
Chemistry: The Central Science · 83 exercises
Problem 78
An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered off, the \(\mathrm{pH}\) is adjusted to about 1 and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and \(\mathrm{H}_{2} \mathrm{~S}\) is again added; no precipitate forms. No precipitate forms upon addition of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). (See Figure 7.23.) The remaining solution shows a yellow color in a flame test (see Figure 7.22). Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: CdS, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{HgO}_{3}, \mathrm{ZnSO}_{4}, \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\), and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) ?
5 step solution
Problem 79
In the course of various qualitative analysis procedures, the following mixtures are encountered: (a) \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cd}^{2+}\), (b) \(\mathrm{Cr}(\mathrm{OH})_{2}\) and \(\mathrm{Fe}(\mathrm{OH})_{3}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{K}^{4}\), (d) \(\mathrm{Ag}^{*}\) and \(\mathrm{Mn}^{2+}\), Suggest how each mixture might be separated.
4 step solution
Problem 80
Suggest how the cations in each of the following solution mixtures can be separated: (a) Na+ and \(\mathrm{Ca}^{2+}\), (b) \(\mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+}\), (c) \(\mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+}\), (d) \(\mathrm{Ag}^{+}\)and \(\mathrm{Hg}^{2+}\).
4 step solution
Problem 82
A student who is in a great hurry to finish his laboratory work decides that his qualitative analysis unknown contains a metal ion from group 4 of Figure 17.23. He therefore tests his sample directly with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) skipping earlier tests for the metal ions in groups 1,2 , and 3. He observes a precipitate and concludes that a metal ion from group 4 is indeed present. Why is this possibly an erroneots conclusion?
4 step solution
Problem 83
Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.
4 step solution
Problem 84
Rainwater is acidic because \(\mathrm{CO}_{2}(\mathrm{~g})\) dissolves in the water, creating carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{5}\). If the rainwater is toe acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, \(\mathrm{CaCO}_{3}\) ). Calculate the concentrations of carbonic acid, bicarbonate ion \(\left(\mathrm{HCO}_{3}{\underline{\phantom{xx}}}^{-}\right)\)and carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) that are in a raindrop that has a \(\mathrm{pH}\) of \(5.60\), assuming that the sum of all three species in the raindrop is \(1.0 \times 10^{-5} \mathrm{M} .\)
4 step solution
Problem 85
Furoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of (a) a solution formed by adding \(25.0 \mathrm{~g}\) of furoic acid and \(30.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.250 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(30.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\) and \(20.0 \mathrm{~mL}\). of \(0.22 \mathrm{M} \mathrm{NaC} \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\) and diluting the total volume to \(125 \mathrm{~mL}\). (c) a solution prepared by adding \(50.0 \mathrm{~mL}\) of \(1.65 \mathrm{M} \mathrm{NaOH}\) solution to \(0.500 \mathrm{~L}\) of \(0.0850 \mathrm{M} \mathrm{HC} \mathrm{H}_{3} \mathrm{O}_{3}\).
9 step solution
Problem 86
The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicater are present in equal concentrations in a solution when the \(\mathrm{pH}\) is \(4.68\). What is the \(\mathrm{p}_{a}\) for bromcresol green?
3 step solution
Problem 87
Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2\). (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and B. (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\) ? (c) What is the value of \(K_{\mathrm{b}}\) for \(\mathrm{B}\) ?
3 step solution
Problem 88
Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(A\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer \(B\) is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in pH for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\). (d) Calculate the change in pH for each buffer upon the addition of \(10 \mathrm{~mL}\). of \(1.00 \mathrm{M} \mathrm{HCl}\).
6 step solution
Problem 89
\mathrm{~A}\( biochemist needs \)750 \mathrm{~mL}\( of an acetic acid-sodium acetate buffer with pH 4.50. Solid sodium acetate ( \)\left.\mathrm{CH}_{3} \mathrm{COONa}\right)\( and glacial acetic acid \)\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\( are available. Glacial acetic acid is \)99 \% \mathrm{CH}_{3} \mathrm{COOH}\( by mass and has a density of \)1.05 \mathrm{~g} / \mathrm{ml}\(. If the buffer is to be \)0.15 \mathrm{M}\( in \)\mathrm{CH}_{3} \mathrm{COOH}\(, how many grams of \)\mathrm{CH}_{4} \mathrm{COONa}$ and how many milliliters of glacial acetic acid must be used?
5 step solution
Problem 90
A sample of \(0.2140 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\). of water and titrated with \(0.0950 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(27.4 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After \(15.0 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 6.50. What is the \(K_{a}\) for the unknown acid?
6 step solution
Problem 91
A sample of \(0.1687 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\). of water and titrated with \(0.1150 \mathrm{M}\) \(\mathrm{NaOH}\). The acid required \(15.5 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After \(7.25 \mathrm{~mL}\) of base had been added in the titration, the pH was found to be \(2.85\). What is the \(K_{a}\) for the unknown acid?
1 step solution
Problem 92
Mathematically prove that the pH at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p}_{u}\) for the acid.
4 step solution
Problem 93
A weak monoprotic acid is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). It requires \(50.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. After \(25.0 \mathrm{~mL}\) of base is added, the pH of the solution is \(3.62\). Estimate the pKa of the weak acid.
5 step solution
Problem 96
How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)
6 step solution
Problem 97
A person suffering from anxiety begins breathing rapidly and as a result suffers alkalosis, an increase in blood \(\mathrm{pH}\). (a) Using Equation 17.10, explain how rapid breathing can cause the \(\mathrm{pH}\) of blood to increase. (b) One cure for this problem is breathing in a paper bag. Why does this procedure lower blood \(\mathrm{pH}\) ?
3 step solution
Problem 99
The solubility of \(\mathrm{CaCO}_{3}\) is \(\mathrm{pH}\) dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{a p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction (c) If we assume that the only sources of \(\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-}\), and \(\mathrm{OH}^{-}\)ions are from the dissolution of \(\mathrm{CaCO}_{3}\), what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)? (e) If the pH is buffered at \(7.5\), what is the molar solubility of \(\mathrm{CaCO}_{3}\) ?
5 step solution
Problem 100
Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{\mathrm{s}}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\), and whose corresponding \(K_{s p}=6.8 \times 10^{-27}\). As discussed in the "Chemistry and Life" box on page 755, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}_{\text {k }}\) whose \(K_{p}=1.0 \times 10^{-40}\), (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.
2 step solution
Problem 102
Calculate the solubility of \(\mathrm{Mg}_{\mathrm{g}}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH} \mathrm{NCl}_{4}\).
5 step solution
Problem 103
The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-4} \mathrm{M}\) for the \(\mathrm{Ba}^{+}\)ion in solution?
3 step solution
Problem 107
The value of \(K_{a p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20}\). The \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{\mathrm{a} 1}=\right.\) \(\left.2.22 ; \mathrm{pK}_{u 2}=6.98 ; \mathrm{p} K_{u 3}=11.50\right)\). When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{p p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\right]\). Why was this a mistake?
6 step solution
Problem 108
The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxe complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{\underline{\phantom{xx}}}^{2-}\), is \(4.6 \times 10^{17}\), What concentration of \(\mathrm{OH}\) is required to dissolve \(0.015 \mathrm{~mol}\) of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?
8 step solution
Problem 109
The value of \(K_{\text {sp }}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is \(2.5 \times 10^{-14}\). (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) ? (b) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}{\underline{\phantom{xx}}}^{2-}\left(K_{f}=5 \times 10^{3}\right)\). If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a \(\mathrm{NaBr}\) solution, what is the initial concentration of \(\mathrm{NaBr}\) needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to \(1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?\)
4 step solution
Problem 112
(a) A 0.1044-g sample of an unknown monoprotic acid requires \(22.10 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{NaOH}\) to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89\). What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.
9 step solution
Problem 113
Aspirin has the structural formula
5 step solution
Problem 114
What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm}\) ? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\).
4 step solution
Problem 115
Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a \(50.00-\mathrm{mL}\) sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M} \mathrm{HQ}\) to reach the end point. Calculate \(K_{\mathrm{wp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\). Compare your result with that in Appendix D. \(25^{\circ} \mathrm{C}\). Suggest a reason for any differences you find between your value and the one in Appendix D.
4 step solution
Problem 116
The osmotic pressure of a saturated solution of strontium sulfate at \(25^{\circ} \mathrm{C}\) is 21 torr. What is the solubility product of this salt at \(25^{\circ} \mathrm{C}\) ?
3 step solution
Problem 117
A concentration of 10-100 parts per billion (by mass) of \(\mathrm{Ag}^{4}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\)can cause adverse health effects. One way to maintain an appropriate concentration of \(\mathrm{Ag}^{+}\)is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix D, calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\)in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) AgI.
3 step solution
Problem 118
Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically the F ion concentration is adjusted to about \(1 \mathrm{ppb}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppb}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)
4 step solution
Problem 119
Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\), which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(\mathrm{~g})\) forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of \(\mathrm{mol} / \mathrm{L}\). (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons). (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to \(350^{\circ} \mathrm{F}\).
3 step solution
Problem 120
In nonaqueous solvents, it is possible to react HF to create \(\mathrm{H}_{2} \mathrm{~F}^{+}\). Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, (d) There is an acid in the nonaqueous medium that is a stronger acid than HF.
5 step solution