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<div class="WordSection1">
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">There are some problems with several of the practice exercises that follow the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Sample Exercises in Ch 16.<span style="mso-spacerun:yes">
</span>The following also has corrections for sample<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">exercises, Go Figure questions and Self-Assessment exercise.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.1:<o:p></o:p></span></b></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></b></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Exercise 16.1:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">It has the wrong answer.
<span style="mso-spacerun:yes"> </span>It asks for the reaction between O2^- and H2O when<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Li2O is put in water.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">LI2O (<span class="SpellE">aq</span>)<span style="mso-spacerun:yes">
</span>+<span style="mso-spacerun:yes"> </span>H2O(</span><i><span style="font-family:"Script MT Bold";mso-bidi-font-family:Calibri">l</span></i><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""> )<span style="mso-spacerun:yes">
</span>-<span class="GramE">><span style="mso-spacerun:yes"> </span>2</span> Li^+ (<span class="SpellE">aq</span>)<span style="mso-spacerun:yes">
</span>+<span style="mso-spacerun:yes"> </span>2 OH- (<span class="SpellE">aq</span>)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">The net ionic
<span class="SpellE">eqn</span> is:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">O^2- (<span class="SpellE">aq</span>)<span style="mso-spacerun:yes">
</span>+<span style="mso-spacerun:yes"> </span>H2O(</span><i><span style="font-family:"Script MT Bold";mso-bidi-font-family:Calibri">l</span></i><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""> )<span style="mso-spacerun:yes">
</span>-<span class="GramE">><span style="mso-spacerun:yes"> </span>2</span> OH- (<span class="SpellE">aq</span>)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.2:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex 16.2:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Has no “Check Answer” button.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Asks for the conjugate acids of HSO3^-, F^-, CH3NH2, PO4^3-.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">This means treat these as bases and add an H+ to get the conj. acid.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">H2SO3, HF, CH3NH3^+, HPO4^2-<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.3:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">This question is asking whether the acid/base reactions lie to the right<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">(mostly product, K>1) or lie to the left (mostly reactant, K<1).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">When you click on the “Check Your Answer” button it gives the answer<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">to Practice Ex. 16.2.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">a) lies to the left (OH-, the stronger base, and stronger acid, H2PO4^-, are<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">on the product side)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>HPO4^2<span class="GramE">-<span style="mso-spacerun:yes"> </span>+</span><span style="mso-spacerun:yes">
</span>H2O(</span><i><span style="font-family:"Script MT Bold";mso-bidi-font-family:Calibri">l</span></i><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""> )<span style="mso-spacerun:yes">
</span>-><span style="mso-spacerun:yes"> </span>H2PO4^-<span style="mso-spacerun:yes">
</span>+<span style="mso-spacerun:yes"> </span>OH-<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>OH- is a stronger base than HPO4^2-<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>H2PO4^- is a stronger acid than H2O<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span><span class="GramE">Actually, once</span> you know which side the stronger acid is located<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>then the stronger base must be on the same side, and vice versa.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>You can answer this one without Figure 16.4 since OH- is the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>strongest base you can have in water.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">b)<span style="mso-spacerun:yes">
</span>lies to the right (OH-, the stronger base, and stronger acid, NH4^+, are<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">on the reactant side)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>NH4^<span class="GramE">+<span style="mso-spacerun:yes"> </span>+</span><span style="mso-spacerun:yes">
</span>OH-<span style="mso-spacerun:yes"> </span>-><span style="mso-spacerun:yes">
</span>NH3<span style="mso-spacerun:yes"> </span>+<span style="mso-spacerun:yes">
</span>H2O<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>OH- is a stronger base than NH3<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>NH4^<span class="GramE">+<span style="mso-spacerun:yes"> </span>is</span> a stronger acid than H2O<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.4:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.5:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">It has the answers for PE 16.3.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">a)<span style="mso-spacerun:yes">
</span>In a sample of lemon juice, [H+] = 3.8 x 10^-4 M.<span style="mso-spacerun:yes">
</span>What is the pH?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>pH = - log(3.8 x 10^-4) = 3.42<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">b) [OH-] = 1.9 x 10^-6
<span class="GramE">M<span style="mso-spacerun:yes"> </span>What</span> is the pH?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>Different ways to do this.<span style="mso-spacerun:yes"> </span>You could use Kw = [H+] [OH-] = 1.0 x 10^-14<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>and find [H+] and the <span class="SpellE">pH.</span><span style="mso-spacerun:yes">
</span>You could use <span class="SpellE">pKw</span> = pH + pOH = 14.00 and<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>solve for pOH and then get pH from this latter eqn.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>pOH = - log [OH-] = - log(1.9 x 10^-6) = 5.72<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>pH = 14.00 – 5.72 = 8.28<span style="mso-spacerun:yes"> </span>(this makes sense since [OH-] > 10^-7 and<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:5">
</span>pH > 7.00 at 25 C means basic).<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Go Figure 16.8:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">It states the solution is pink when phenolphthalein is added and asks what it<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">tells you about the
<span class="SpellE">pH.</span><span style="mso-spacerun:yes"> </span>It then states pH > 10.<span style="mso-spacerun:yes">
</span><span class="GramE">Actually, it</span> would mean pH >8.2<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">since it turns a light pink around a pH = 8.2.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.6:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">A solution formed by dissolving an antacid table has a pOH of 4.82. What is<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">the [H+] in this solution?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">It gives the answer for PE 16.5.<span style="mso-spacerun:yes">
</span>The correct answer is:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>pH = 14.00 – 4.82 = 9.18<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>[H+] = 10^(-pH) = 10^-(9.18) = 6.6 x 10^-10<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.5:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.7:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">An aqueous solution of HNO3 had a pH of 2.34.<span style="mso-spacerun:yes">
</span>What is the conc. of the acid?<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">This has the answer to PE 16.6.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">HNO3 is a strong acid so it completely ionizes,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>HNO3 -> H<span class="GramE">+<span style="mso-spacerun:yes"> </span>+</span><span style="mso-spacerun:yes">
</span>NO3^-<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">There’s essentially no HNO3 at equilibrium.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>[H+]_eq = [HNO3]_original<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>[H+] = 10^-(pH) = 10^-(2.34) = 4.57 x 10^-3 M = 4.6 x 10^-3 M<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><span style="mso-tab-count:1">
</span>Thus, [HNO3]_original = 4.6 x 10^-3 M<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Sample Ex. 16.8:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Part (b) Method 1:<span style="mso-spacerun:yes">
</span>the [OH-] in the denominator should be 0.0022 M.<span style="mso-spacerun:yes">
</span>The<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">answer given for [H+] is correct.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.6:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.9:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Just incorrectly rounded. They have the answer as 1.5 x 10^-5.<span style="mso-spacerun:yes">
</span>I get<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">1.552 x 10^-5 M when carrying extra digits and this rounds 1.6 x 10^-5 M.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Go Figure 16.11:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">The explanation given may be hard to understand.<span style="mso-spacerun:yes">
</span>I went over this in lecture.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Can be explained by thinking about diluting the solution.<span style="mso-spacerun:yes">
</span>When the volume of<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">solution is increased by adding H2O all the conc. of reactants and products<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">decrease (<span class="GramE">i.e.</span> total molarities in solution decreases).<span style="mso-spacerun:yes">
</span>When that happens, the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">reaction shifts to the right, toward more moles in solution to raise the total<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">moles (molarities), just like for gases when P (and molarity) is decreased by<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">increasing the volume, the rxn shifts to increase the moles of gas in solution.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">In solution, increasing the volume of solution by adding solvent the rxn shifts<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">to
<span class="SpellE">inc</span> the moles of solutes in solution.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Self-Assessment 16.16:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-fareast-font-family:"Times New Roman";color:black">Rounded incorrectly.<span style="mso-spacerun:yes">
</span>I get Ka = 2.579 x 10<sup>-3</sup> which rounds to 2.6 x 10<sup>-3</sup><o:p></o:p></span></p>
<p class="MsoNormal"><span style="mso-fareast-font-family:"Times New Roman";mso-bidi-font-family:Calibri;color:black"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.7:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Sample Exercise 16.13:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Uses rounded value of OH- to determine pOH and
<span class="SpellE">pH.</span><span style="mso-spacerun:yes"> </span>You should not<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">round in intermediate steps.<span style="mso-spacerun:yes">
</span>You should carry at least one extra digit than<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">the number of sig. fig. into subsequent steps.<span style="mso-spacerun:yes">
</span>If you don’t round at all<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">you will get pH = 11.22.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Section 16.9:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Practice Ex. 16.17:<o:p></o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">The answer of acidic is correct.<span style="mso-spacerun:yes">
</span>The reason is the Ka for HC6H5O7^2-<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">acting as an acid is bigger than the
<span class="SpellE">Kb</span> for it acting as a base.<o:p></o:p></span></p>
<p class="MsoNormal"><b><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></b></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman""><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif;mso-bidi-font-family:"Times New Roman"">Dr. Zellmer<o:p></o:p></span></p>
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