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only) paper 1 questions</a></label></li></ul></div> <button id="show-periodic-table" class="btn btn-default btn-block" style="margin-bottom: 10px"><i class="fa fa-table"></i> Periodic table</button> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Acids and Bases AHL (HL only) paper 1 questions <a href="#" class="mark-page-favorite pull-right" data-pid="2761" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../2075/paper-1-exam-questions.html">Paper 1 Exam Questions</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Acids and Bases AHL (HL only) paper 1 questions</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><strong>Topics 18.1 to 18.3</strong></p> <p><strong>Paper 1 style questions </strong>are multiple choice. You are <strong>not permitted to use a calculator or the data book</strong> for these questions, but you should use a periodic table.</p> <p>A <strong>periodic table pop-up</strong> is available on the left hand menu.</p> <div class="greenBg"> <div class="tib-quiz" data-stats="7-878-2761"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which combination will produce an alkaline buffer solution in water?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0.10 mol CH<sub>3</sub>COOH and 0.10 mol CH<sub>3</sub>COONa</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.10 mol CH<sub>3</sub>COOH 0.20 mol NaOH</span></label> </p><p><label class="radio"> <input type="radio"> 0.20 mol NH<sub>3</sub> and 0.10 mol H<sub>2</sub>SO<sub>4</sub> </label></p><p><label class="radio"> <input class="c" type="radio"> 0.20 mol NH<sub>3</sub> and 0.05 mol H<sub>2</sub>SO<sub>4</sub> </label></p></div><div class="q-explanation"><p>An alkaline buffer solution is the solution of a weak base and the salt of the weak base (that <strong>resists/opposes </strong>changes in pH upon addition of small amounts of acid or base).</p><p>0.20 mol NH<sub>3</sub> and 0.05 mol H<sub>2</sub>SO<sub>4</sub> will produce an alkaline buffer, since 0.05 mol H<sub>2</sub>SO<sub>4</sub> will neutralise half (0.10 mol) of the NH<sub>3</sub> (1:2 ratio), leaving unreacted weak base, NH<sub>3</sub>, and the salt of the weak base, (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub>.</p><p style="text-align: center;">2NH<sub>3</sub> + H<sub>2</sub>SO<sub>4</sub> → (NH<sub>4</sub>)<sub>2</sub>SO<sub>4</sub></p><p><strong>0.20 mol NH<sub>3</sub> and 0.05 mol H<sub>2</sub>SO<sub>4</sub> </strong>is therefore the correct answer.</p><p><strong>Incorrect answers</strong></p><p>0.20 mol NH<sub>3</sub> and 0.10 mol H<sub>2</sub>SO<sub>4</sub> will result in <strong>all </strong>of the ammonia, NH<sub>3</sub>, being neutralised (1:2 ratio) so this cannot form a buffer.</p><p>A combination of ethanoic acid, CH<sub>3</sub>COOH, and sodium ethanoate, CH<sub>3</sub>COONa, (in any reasonable quantities) will form a buffer solution, but that will be an <strong>acidic </strong>buffer.</p><p>A combination of <strong>an excess of ethanoic acid</strong> with sodium hydroxide will also form a buffer solution, since all of the sodium hydroxide will be neutralised and will produce sodium ethanoate, but <strong>an excess of NaOH</strong> will neutralise all the acid and the solution will not be a buffer (as is the case in this question).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following could correctly be described as a nucleophile?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Bronsted-Lowry acid</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Lewis base</span></label> </p><p><label class="radio"> <input type="radio"> <span>Bronsted-Lowry base</span></label> </p><p><label class="radio"> <input type="radio"> <span>Lewis acid</span></label> </p></div><div class="q-explanation"><p>A nucleophile is a substance which can donate a lone pair of electrons, which is an identical description to that of a Lewis base.</p><p>Likewise, an electrophile is equivalent to a Lewis acid as it can accept a lone pair of electrons.</p><p><strong>Lewis base</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which compound is alkaline when dissolved in water?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>NH<sub>4</sub>Cl</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>CH<sub>3</sub>CO<sub>2</sub>Na</span></label> </p><p><label class="radio"> <input type="radio"> <span>NaCl</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>COOH</span></label> </p></div><div class="q-explanation"><p>CH<sub>3</sub>COOH is an acid.</p><p>The other three options are all salts.</p><p>As a useful rule-of-thumb, salts formed from strong acids and strong bases (e.g. NaCl) and those from weak acids and weak bases give approximately neutral solutions of pH=7.</p><p>Salts formed from strong acids and weak bases (e.g. NH<sub>4</sub>Cl) are acidic and those from weak acids and strong bases (e.g. CH<sub>3</sub>COONa) are alkaline.</p><p>Salt hydrolysis, is due to the reaction of the conjugate bases/acids<strong> </strong>(salt ions) of <strong>weak </strong>acids/bases with <strong>water</strong>:</p><p><em>E.g. NH<sub>4</sub><sup>+</sup> + H<sub>2</sub>O </em><em>⇌</em><em> NH<sub>4</sub>OH + <strong>H<sup>+ </sup></strong></em></p><p><em>E.g. CH<sub>3</sub>COO</em><em><sup>–</sup></em> <em>+ H<sub>2</sub>O </em><em>⇌</em><em> CH<sub>3</sub>COOH + <strong>OH</strong></em><strong><em><sup>–</sup></em></strong></p><p><strong>CH<sub>3</sub>CO<sub>2</sub>Na</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which can act as a Lewis acid, but not a Bronsted-Lowry acid?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>NH<sub>4</sub><sup>+</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>HCl</span></label> </p><p><label class="radio"> <input type="radio"> <span>NF<sub>3</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> BF<sub>3</sub> </label></p></div><div class="q-explanation"><p>A Lewis acid is a substance that can accept a pair of electrons from a Lewis base, forming a co-ordinate (dative) bond.</p><p>A Bronsted-Lowry acid can donate an H<sup>+</sup> ion.</p><p>Boron in BF<sub>3</sub> is electron deficient (it has an incomplete octet) and so it can accept a pair of electrons. BF<sub>3</sub> cannot act as a Bronsted-Lowry acid, as it has no hydrogens.</p><p>None of the other species can act as Lewis acids, as none of the atoms are electron deficient, and they all have complete octets (and cannot expand them); or two electrons in the case of hydrogen.</p><p>The correct answer is therefore <strong>BF<sub>3</sub></strong>.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator:</em></p><p>Which solution, of equal concentrations, will have the lowest pH?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> CH<sub>2</sub>ClCOOH (Ka = 1.4 × 10<sup>−3</sup>) </label></p><p><label class="radio"> <input type="radio"> <span>C<sub>6</sub>H<sub>5</sub>OH (Ka = 1.0 × 10<sup>−10</sup>)</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>COOH (pKa = 4.76)</span></label> </p><p><label class="radio"> <input type="radio"> <span>CH<sub>3</sub>CH<sub>2</sub>COOH (pKa = 4.87)</span></label> </p></div><div class="q-explanation"><p>Weak acids only partially dissociate: HA ⇌ H<sup>+</sup> + A<sup>−</sup></p><p><span class="math-tex">\(K_a = {{ [A^-][H^+] } \over [HA]}\)</span></p><p>Therefore the Ka for a stronger acid will be greater than for a weaker acid (more ions on top of fraction).</p><p>pKa = −log<sub>10</sub>Ka</p><p>Therefore higher Ka leads to <strong>lower </strong>pKa (similarly to the relationship between [H<sup>+</sup>] and the pH scale).</p><p>Stronger acids will always have lower pH than weak acids of the same concentration because of the increased dissociation.</p><p>Without a calculator we need to remember that pKa is a logarithmic scale, so Ka = 1.4 × 10<sup>−3</sup> will give a <strong>pKa of approx. 3</strong> (the × 10<sup>−3</sup> tells us this) and Ka = 1.0 × 10<sup>−10</sup> will give a <strong>pKa of approx.10</strong>.</p><p>The strongest acid (with a pKa of approx. 3) is therefore CH<sub>2</sub>ClCOOH, and so this will have the lowest pH.</p><p>Therefore <strong>CH<sub>2</sub>ClCOOH (Ka = 1.4 × 10<sup>−3</sup>)</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph below represents a pH titration curve. Which statements are correct?</p><p>1: The pH curve could have been produced by titration of ethanoic acid with sodium hydroxide solution.</p><p>2: The equivalence point is at pH 7.</p><p>3: The buffer region includes the half-equivalence point (12.5cm<sup>3</sup> added).</p><p><img alt="" src="../../images/acids-and-bases/wvss.png" style="width: 240px; height: 226px;"></p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>1 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p></div><div class="q-explanation"><p>The shapes of the four pH titration curves need to be learned.</p><p>The starting pH (∼3), high finishing pH and the steep inflection in the curve above pH 7 indicate that this is a weak acid, strong base pH curve. Therefore the curve could have been produced by titration of ethanoic acid with sodium hydroxide solution.</p><p>The <strong>equivalence point</strong> is the point at which the moles of acid = moles of base, and for a weak acid and strong base pH curve, this is typically around<strong> pH 9 (not pH 7).</strong></p><p>At half the equivalence point (at 12.5cm<sup>3</sup>) half of the acid has been neutralised so the concentration of acid and salt is the same: [HA] = [A<sup>–</sup>] which means this is a buffer solution, and so this part of the curve is known as the buffer region.</p><p><strong>1 and 3 only </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>What changes occur in water as temperature increases (Kw = 1 × 10<sup>−14</sup> at 298K)?</p><p>H<sub>2</sub>O (l) ⇌ H<sup>+</sup> (aq) + OH<sup>−</sup> (aq) ΔH = +56 kJ</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> Kw will decrease and the pH will increase</label></p><p><label class="radio"><input type="radio"> Kw will increase and the pH will increase</label></p><p><label class="radio"><input type="radio"> Kw will decrease and the pH will decrease</label></p><p><label class="radio"><input class="c" type="radio"> Kw will increase and the pH will decrease</label></p></div><div class="q-explanation"><p>If the temperature is increased the equation will shift in the endothermic direction, which means the equilibrium will shift to the right.</p><p>Shifting the equilibrium to the right will increase the quantity of H<sup>+</sup> ions and OH<sup>−</sup> ions in the aqueous solution; [H<sup>+</sup>] and [OH<sup>−</sup>] will increase.</p><p>Kw = [H<sup>+</sup>] × [OH<sup>−</sup>]</p><p>Kw will therefore<strong> increase</strong> as the concentration of H<sup>+</sup> and OH<sup>−</sup> ions increases, and pH will <strong>decrease</strong> as H<sup>+</sup> ion concentration increases, since pH=−log<sub>10</sub>[H<sup>+</sup>].</p><p><strong>Kw will increase and the pH will decrease</strong> is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>An indicator, HIn, has pKa = 4.2.</p><p style="text-align: center;">HIn (aq) ⇌ H<sup>+</sup> (aq) + In<sup>−</sup> (aq)</p><p>HIn (aq) is yellow, and In<sup>−</sup> (aq) is blue.</p><p>Which statement/s are correct?</p><p>1: At pH 7.0 the indicator would be yellow</p><p>2: At pH 4.2 the indicator would be green</p><p>3: At pH 4.2, [Hin] = [In<sup>−</sup>]</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation"><p>Indicators are weak acids that change colour when [HIn] = [In<sup>−</sup>] (In practice the colour changes across a pH range). At the point when [HIn] = [In<sup>−</sup>], K<sub>a</sub> = [H<sup>+</sup>] and pH = pKa (see expression below; [HIn] and [In<sup>−</sup>] cancel each other out).</p><p><span class="math-tex">\(K_a = {{[H^+][In^-]} \over [HIn]}\)</span></p><p>Therefore indicators change colour across a range 'straddling' pH = pKa.</p><p>Therefore, for the indicator in this question, at <strong>pH 4.2, [Hin] = [In<sup>−</sup>]</strong> and since the protonated and deprotonated forms are yellow and blue, <strong>a 50:50 mixture will be green</strong>.</p><p>At pH 7.0 the indicator in this question will be blue (not yellow), since less H+ will push the equilibrium right and [Hin] << [In<sup>−</sup>]</p><p><strong>2 and 3 only </strong>is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which combination of acid and base is most likely to have a pH of 5.5 at the equivalence point in a titration?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Propanoic acid and potassium hydroxide</span></label> </p><p><label class="radio"> <input type="radio"> <span>Hydrochloric acid and potassium hydroxide</span></label> </p><p><label class="radio"> <input type="radio"> <span>Propanoic acid and ammonia solution</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Hydrochloric acid and ammonia solution</span></label> </p></div><div class="q-explanation"><p>The shapes of the four pH titration curves need to be learned.</p><p><img alt="" src="../../images/acids-and-bases/svsw.png" style="width: 240px; height: 225px;"></p><p>This is a strong acid, weak base pH curve. Notice the low starting pH, finishing pH of ∼11, and the steep inflection in the curve below pH 7.</p><p>The equivalence point is at pH 5/6.</p><p><strong>Hydrochloric acid and ammonia solution</strong><strong> </strong>(strong acid and weak base) is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Complete this question without a calculator: </em>A buffer solution can be produced by mixing ethanoic acid, CH<sub>3</sub>COOH (aq), and sodium hydroxide, NaOH (aq).</p><p>If 25.0 cm<sup>3</sup> of 0.100 mol dm<sup>−3</sup> CH<sub>3</sub>COOH (aq) is used, what volume of 0.100 mol dm<sup>−3</sup> NaOH (aq) will produce a buffer, and what will the pH of the buffer be?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>12.5 cm<sup>3</sup> of NaOH and pH = 9.2</span></label> </p><p><label class="radio"> <input type="radio"> <span>50.0 cm<sup>3</sup> of NaOH and pH = 4.8</span></label> </p><p><label class="radio"> <input type="radio"> 50.0 cm<sup>3</sup> of NaOH and pH = 9.2 </label></p><p><label class="radio"> <input class="c" type="radio"> 12.5 cm<sup>3</sup> of NaOH and pH = 4.8 </label></p></div><div class="q-explanation"><p>A combination of ethanoic acid, CH<sub>3</sub>COOH, and the salt of the weak acid e.g. sodium ethanoate, CH<sub>3</sub>COONa, (in any reasonable quantities) will form a buffer solution, and that will be an <strong>acidic </strong>buffer (with a pH < 7).</p><p>A combination of <strong>an excess of ethanoic acid</strong> with sodium hydroxide will also form a buffer solution, since all of the sodium hydroxide will be neutralised and will produce sodium ethanoate (CH<sub>3</sub>COOH + NaOH → CH<sub>3</sub>COONa + H<sub>2</sub>O), and this results in a mixture of unreacted ethanoic acid and sodium ethanoate.</p><p><strong>12.5 cm<sup>3</sup> of NaOH and pH = 4.8 </strong>is therefore the correct answer, since this volume of NaOH will neutralise half of the ethanoic acid, and the resulting buffer will be acidic (pH < 7).</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> <div class="page-container panel-self-assessment" data-id="2761"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Acids and Bases AHL (HL only) paper 1 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