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	<title>DP Chemistry: Misuse of formulas</title>
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fusion & nuclear fission reactions</a></li><li><a href="../19429/solar-energy-1.html">Solar energy</a></li><li><a href="../19437/environmental-impact-global-warming-1.html">Environmental impact - global warming</a></li><li><a href="../19453/electrochemistry-rechargeable-batteries-fuel-cells-1.html">Electrochemistry, rechargeable batteries & fuel cells</a></li><li><a href="../19458/nuclear-fusion-fission-hl--1.html">Nuclear fusion & fission (HL)    </a></li><li><a href="../19470/photovoltaic-cells-and-dsscs-1.html">Photovoltaic cells and DSSCs</a></li></ul></li><li><a href="../19078/option-d-1.html" class="father">Option D</a><ul class="level-2"><li><a href="../19079/pharmaceutical-products-drug-action--1.html">Pharmaceutical products & drug action </a></li><li><a href="../19158/aspirin-penicillin--1.html">Aspirin & penicillin  </a></li><li><a href="../19174/opiates-1.html">Opiates</a></li><li><a href="../19184/ph-regulation-of-the-stomach-1.html">pH regulation of the stomach</a></li><li><a href="../19194/antiviral-medications--1.html">Antiviral medications </a></li><li><a href="../19211/environmental-impact-of-some-medications-1.html">Environmental impact of some medications</a></li><li><a href="../19254/taxol-a-chiral-auxiliary-case-study-1.html">Taxol - a chiral auxiliary case study</a></li><li><a href="../19258/nuclear-medicine--1.html">Nuclear medicine  </a></li><li><a href="../19273/drug-detection-analysis--1.html">Drug detection & analysis </a></li></ul></li></ul></li><li><a href="../16592/practical-scheme-of-work-ia--1.html">Practical scheme of work & IA </a><ul class="level-1"><li><a href="../16682/internal-assessment-1.html" class="father">Internal Assessment</a><ul class="level-2"><li><a href="../18892/scaffolding-the-investigation-1.html">&#039;Scaffolding&#039; the investigation</a></li><li><a href="../18896/timing-organisation--1.html">Timing & organisation </a></li><li><a href="../18905/choosing-the-research-question-1.html">Choosing the research question</a></li><li><a href="../18946/personal-engagement-1.html">Personal engagement</a></li><li><a href="../18950/exploration-1.html">Exploration</a></li><li><a href="../18957/analysis-1.html">Analysis</a></li><li><a href="../18965/evaluation-1.html">Evaluation</a></li><li><a href="../18966/communication-1.html">Communication</a></li><li><a href="../19882/internal-standardization-of-the-ia-1.html">Internal standardization of the IA</a></li><li><a href="../19901/submitting-the-samples-for-moderation-2.html">Submitting the samples for moderation</a></li><li><a href="../33754/ten-suggestions-for-ias-using-secondary-data-1.html">Ten suggestions for IAs using secondary data</a></li><li><a href="../37544/gaining-full-marks-for-a-databased-ia--1.html">Gaining full marks for a databased IA </a></li><li><a href="../19506/ia-example-marking-exercise--1.html">IA example & marking exercise </a></li><li><a href="../23929/genuine-examples-of-moderated-ia-reports-1.html">Genuine examples of moderated IA reports</a></li><li><a href="../25234/examples-of-teacher-marked-ia-reports--1.html">Examples of teacher-marked IA reports </a></li><li><a href="../37542/history-of-internal-assessment-1.html">History of Internal Assessment</a></li></ul></li><li><a href="../16598/mandatory-laboratory-components-1.html" class="father">Mandatory laboratory components</a><ul class="level-2"><li><a href="../16613/formula-of-magnesium-oxide--1.html">Formula of magnesium oxide </a></li><li><a href="../16612/determining-the-mr-of-an-unknown-gas-1.html">Determining the <i>M</i><sub>r</sub> of an unknown gas</a></li><li><a href="../16615/acid-base-titrations--1.html">Acid-base titrations </a></li><li><a href="../16616/a-green-acid-base-practical-1.html">A green acid-base practical</a></li><li><a href="../16617/analysis-of-aspirin-tablets-1.html">Analysis of aspirin tablets</a></li><li><a href="../16618/caco3-in-egg-shells-1.html">CaCO<sub>3</sub> in egg shells</a></li><li><a href="../16644/enthalpy-changes-1.html">Enthalpy changes</a></li><li><a href="../16631/reaction-rates-1.html">Reaction rates</a></li><li><a href="../16635/rate-dependent-factors-1.html">Rate-dependent factors</a></li><li><a href="../16636/determining-ea-for-a-reaction-1.html">Determining <i>E</i><sub>a</sub> for a reaction</a></li><li><a href="../16637/iodination-of-propanone-1.html">Iodination of propanone</a></li><li><a href="../18144/titrations-with-a-ph-meter-1.html">Titrations with a pH meter</a></li><li><a href="../18355/redox-titration-with-kmno4--1.html">Redox titration with KMnO<sub>4</sub> </a></li><li><a href="../16640/voltaic-cells-1.html">Voltaic cells</a></li><li><a href="../16659/3-d-molecular-modelling--1.html">3-D molecular modelling </a></li></ul></li><li><a href="../17168/other-good-practicals-1.html" class="father">Other good practicals</a><ul class="level-2"><li><a href="../17196/common-chemical-reactions-1.html">Common chemical reactions</a></li><li><a href="../227/elements-oxides-of-the-third-period--1.html">Elements & oxides of the third period </a></li><li><a href="../17239/the-halogens-1.html">The halogens</a></li><li><a href="../17162/boiling-points-of-mixtures-1.html">Boiling points of mixtures</a></li><li><a href="../17170/polarity-of-molecules-1.html">Polarity of molecules</a></li><li><a href="../18083/le-chateliers-principle--1.html">Le Chatelier&#039;s principle </a></li><li><a href="../19582/determining-kc-for-an-esterification-reaction-1.html">Determining <i>K</i><sub>c</sub> for an esterification reaction</a></li><li><a href="../18164/redox-reactions-of-vanadium--1.html">Redox reactions of vanadium </a></li><li><a href="../18351/chlorine-in-swimming-pools--1.html">Chlorine in swimming pools </a></li><li><a href="../18178/analysis-of-cuii-ions-in-solution--1.html">Analysis of Cu(II) ions in solution </a></li><li><a href="../18179/percentage-of-copper-in-brass-1.html">Percentage of copper in brass</a></li><li><a href="../18356/electrolytic-cells--1.html">Electrolytic cells </a></li><li><a href="../18598/reactions-of-organic-compounds-1.html">Reactions of organic compounds</a></li><li><a href="../18752/hydrolysis-of-halogenoalkanes-1.html">Hydrolysis of halogenoalkanes</a></li><li><a href="../18784/preparation-of-13-dinitrobenzene--1.html">Preparation of 1,3-dinitrobenzene </a></li><li><a href="../19022/determination-of-an-organic-structure-1.html">Determination of an organic structure</a></li><li><a href="../19887/preparation-of-nylon-66-1.html">Preparation of nylon 6,6</a></li><li><a href="../19343/determination-of-vitamin-c-content--1.html">Determination of vitamin C content </a></li><li><a href="../19342/hydrolysis-of-starch--1.html">Hydrolysis of starch </a></li><li><a href="../19159/preparation-purification-of-aspirin--1.html">Preparation & purification of aspirin </a></li></ul></li><li><a href="../18851/ict-in-practical-work-1.html" class="father">ICT in practical work</a><ul class="level-2"><li><a href="../18852/data-logging--1.html">Data logging </a></li><li><a href="../18853/software-for-graph-plotting--1.html">Software for graph plotting </a></li><li><a href="../18854/spreadsheets-for-data-processing--1.html">Spreadsheets for data processing </a></li><li><a href="../18855/databases-1.html">Databases</a></li><li><a href="../18856/computer-modelling-simulations--1.html">Computer modelling & simulations </a></li></ul></li><li><a href="../16594/group-4-project-1.html" class="father">Group 4 Project</a><ul class="level-2"><li><a href="../16595/practicalities-1.html">Practicalities</a></li><li><a href="../16596/reflective-statement--2.html">Reflective statement </a></li></ul></li><li><a href="../16666/a-new-laboratory-1.html">A new laboratory?</a></li></ul></li><li class="selected"><a href="../16284/exams-1.html">Exams</a><ul class="level-1"><li><a href="../16286/essential-facts-1.html" class="father">Essential Facts</a><ul class="level-2"><li><a href="../16287/objectives-command-terms-1.html">Objectives & command terms</a></li><li><a href="../16288/grade-descriptors-1.html">Grade descriptors</a></li><li><a href="../16289/grade-boundaries-1.html">Grade boundaries</a></li></ul></li><li><a href="../16302/paper-1-multiple-choice-1.html" class="father">Paper 1 (Multiple Choice)</a><ul class="level-2"><li><a href="../16303/advice-to-students-paper-1-1.html">Advice to students (Paper 1)</a></li><li><a href="../31515/practice-paper-1-exams-1.html">Practice Paper 1 exams</a></li></ul></li><li class="selected"><a href="../16305/paper-2-1.html" class="father">Paper 2</a><ul class="level-2"><li><a href="../16306/advice-to-students-paper-2-1.html">Advice to students (Paper 2)</a></li><li class="selected"><a href="../16307/areas-of-difficulty-1.html">Areas of difficulty</a></li></ul></li><li><a href="../16313/paper-3--1.html" class="father">Paper 3 </a><ul class="level-2"><li><a href="../16314/advice-to-students-paper-3-1.html">Advice to 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href="../16325/faqs-1.html">FAQs</a></li></ul></li><li><a href="../3147/ib-core-1.html">IB Core</a><ul class="level-1"><li><a href="../3149/learner-profile-1.html">Learner Profile</a></li><li><a href="../21735/extended-essays--1.html" class="father">Extended Essays </a><ul class="level-2"><li><a href="../21736/overview--1.html">Overview </a></li><li><a href="../21780/responsibilities-1.html">Responsibilities</a></li><li><a href="../21790/resources-1.html">Resources</a></li><li><a href="../21800/research-1.html">Research</a></li><li><a href="../21804/writing-the-essay-1.html">Writing the essay</a></li><li><a href="../21810/assessment-1.html">Assessment</a></li><li><a href="../22434/summary-of-advice--1.html">Summary of advice </a></li><li><a href="../25272/faqs--1.html">FAQs </a></li></ul></li><li><a href="../307/theory-of-knowledge-tok-1.html" class="father">Theory of Knowledge (TOK)</a><ul class="level-2"><li><a href="../41171/the-12-tok-concepts-1.html">The 12 TOK concepts</a></li><li><a href="../3507/why-tok-chemistry-1.html">Why TOK & Chemistry?</a></li><li><a href="../3505/a-modern-paradigm-1.html">A modern paradigm</a></li></ul></li></ul></li><li><a href="../17783/fast-track-to-tests-questions-1.html">Fast track to tests & questions</a><ul class="level-1"><li><a href="../17784/sl-multiple-choice-tests-on-individual-topics-1.html">SL Multiple choice tests on individual topics</a></li><li><a href="../24351/sl-multiple-choice-quizzes-on-sub-topics-1.html">SL Multiple choice &#039;quizzes&#039; on sub-topics</a></li><li><a href="../31519/sl-practice-paper-1-exams-1.html">SL Practice Paper 1 exams</a></li><li><a href="../17786/sl-questions-on-each-sub-topic-1.html">SL Questions on each sub-topic</a></li><li><a href="../20431/sl-paper-3-section-a-questions-1.html">SL Paper 3 Section A questions</a></li><li><a href="../17792/sl-questions-on-the-options-1.html">SL Questions on the options</a></li><li><a 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</div><section id="main-content"><div id="toc-1672972255" class="toc"><a href="#" class="toc-switcher" title="Show table of contents"><small><i class="fa fa-reorder"></i> Table of contents <span class="pull-right"><i class="fa fa-chevron-down"></i></span></small></a></div><p>Students often use mathematical formulas to solve chemical problems in exams and sometimes arrive at the wrong answer. This page looks at why this is so and encourages students to work always from first principles rather than mindlessly &#39;plug&#39; values into a formula that they are unable to derive for themselves.&nbsp; It gives worked examples and provides some further practice questions with worked answers.</p><p style="text-align: center;"><img alt="" class="noborder" src="../../../ib/chemistry/images/2014%20Exams/m1v1%3Dm2v2.png" style="width: 234px; height: 88px;"></p><div class="blueBg"><h1>Introduction</h1><p>Both mathematics and science make much use of formulas to solve problems. There is nothing inherently wrong in using a formula. In fact in many cases it can save a considerable amount of time. For example, Einstein is reputed to have said that compound interest is the eighth wonder of the world &ndash; &ldquo;He who understands it earns it&hellip; he who doesn&rsquo;t&hellip; pays it.&rdquo;. A person who 30 years ago put $1000 dollars in an investment account that paid 7% interest per year ($0.07 per dollar) would have gained 30 x $70 in interest if the interest had been withdrawn at the end of each year giving a total sum of $3100 if the original investment is also included. However if the interest had been left in the account the formula A = P (1 + r)<sup>t</sup> (where A = the future value, P = the principal investment amount, r = the interest rate and t = time) shows that the $1000 initial investment would have grown to $7612 after 30 years. If it had been left for 50 years the figures would have been $4500 and $29460 respectively. Clearly using this particular formula saves all the tedious work of working out the increase and compounding it for each year etc.</p><hr class="hidden"><p>Unfortunately some chemistry students who make considerable use of formulas in exams often arrive at the wrong answer because they have either misused a formula or have been unable to apply the correct formula to solve the problem.</p><hr class="hidden"><p>There are several reasons why this may be so.</p><hr class="hidden"></div><hr class="hidden"><div class="yellowBg"><h1>Reasons for errors using formulas</h1><h4>1. The formula is remembered wrongly.</h4><p>A simple formula often used to find the concentration of a solution is:</p><p>c = n / V (where c is the concentration, n is &lsquo;number of moles&rsquo; (actually amount in mol) and V is the volume).</p><p>I&rsquo;ve seen several variations of this in exams, e.g. c = V / n or n = c / V. The key to stating formulas correctly is to check the units. Concentration is normally expressed as mol dm<sup>&minus;3</sup>. Once a student realises this the formula is redundant as mol dm<sup>&minus;</sup><sup>3</sup> literally means the amount in mol divided by the volume in litres (dm<sup>3</sup>). Another example where the formula is often quoted wrongly is the relationship between the velocity of light (<em>c</em>),&nbsp; the frequency (&nu;) and the wavelength (&lambda;).</p><p>Looking at the units it is obvious that velocity (m s<sup>&minus;1</sup>) = wavelength (m) x frequency (s<sup>&minus;1</sup>), i.e. <em>c</em> = <em>&lambda;&nu;</em>.</p><hr class="hidden"><hr class="hidden"><h4>2. The formula is applied wrongly</h4><p>One of the most common examples of this is the formula M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> which students often use to solve titration problems. It does work in the simplest case when one mole of an acid is neutralised by one mole of a base,</p><p>e.g. HCl(aq) + NaOH(aq) &rarr; NaCl(aq) + H<sub>2</sub>O(l)</p><p>but certainly does not work when it comes to the neutralisation of diprotic acids,</p><p>e.g. H<sub>2</sub>SO<sub>4</sub>(aq) + 2NaOH(aq) &rarr; Na<sub>2</sub>SO<sub>4</sub>(aq) + 2H<sub>2</sub>O(l).</p><p>My advice is to never use it! Instead always work from first principles.</p><h4><span style="color:#FF0000;">Worked example:</span></h4><p>26.80 cm<sup>3</sup> of a solution of sulfuric acid was required to neutralise 25.00 cm<sup>3</sup> of 0.150 mol dm<sup>&minus;3</sup> potassium hydroxide solution. Calculate the concentration of the sulfuric acid solution in mol dm<sup>-3</sup>.</p><p><strong>Step 1: </strong>Write the stoichiometric equation</p><p>H<sub>2</sub>SO<sub>4</sub>(aq) + 2KOH(aq) &rarr; K<sub>2</sub>SO<sub>4</sub>(aq) + 2H<sub>2</sub>O(l)</p><hr class="hidden"><p><strong>Step 2: </strong>Calculate the amount (in mol) of potassium hydroxide present.</p><p>1000 cm<sup>3</sup> of 0.150 mol&nbsp;dm<sup>&minus;3</sup> KOH contains 0.150 mol</p><p>so 25.00 cm<sup>3</sup> contains (25.00/1000) x 0.150 = 3.75 x 10<sup>&minus;3</sup> mol.</p><hr class="hidden"><p><strong>Step 3: </strong>Calculate the amount (in mol) of sulfuric acid required to neutralise the potassium hydroxide solution.</p><p>From the equation 1 mol of H<sub>2</sub>SO<sub>4</sub> neutralises 2 mol of KOH so amount of H<sub>2</sub>SO<sub>4</sub> required = &frac12; x 3.75 x&nbsp;10<sup>&minus;3</sup> = 1.875 x&nbsp;10<sup>&minus;3</sup> mol.</p><hr class="hidden"><p><strong>Step 4: </strong>Calculate the concentration of the sulfuric acid in mol dm<sup>&minus;3</sup>.</p><p>26.80 cm<sup>3</sup> of H<sub>2</sub>SO<sub>4</sub> contains 1.875 x&nbsp;10<sup>&minus;3</sup> mol</p><p>1000 cm<sup>3</sup> of H<sub>2</sub>SO<sub>4</sub> contains (1000/26.80) x 1.875 x&nbsp;10<sup>&minus;3</sup> = 0.0700 mol</p><p>so the concentration of the sulfuric acid = 0.0700 mol dm<sup>&minus;3</sup> (to 3 significant figures).</p><hr class="hidden"><hr class="hidden"><h4>3. A formula cannot be used at all.</h4><p>A back to basics stepwise approach is the only way to answer titration questions where a solid, rather than a solution, is being titrated as the formula M<sub>1</sub>V<sub>1</sub> = M<sub>2</sub>V<sub>2</sub> cannot be applied.</p><h4><span style="color:#FF0000;">Example:</span></h4><p>What mass of calcium carbonate will react exactly with 20.00 cm<sup>3</sup> of 0.200 mol&nbsp;dm<sup>&minus;3</sup> hydrochloric acid?</p><p>Equation: 2HCl(aq) + CaCO<sub>3</sub>(s) &rarr; CaCl<sub>2</sub>(aq) + CO<sub>2</sub>(g) + H<sub>2</sub>O(l)</p><p>Amount of acid reacting = 20.00/1000 x 0.200 = 4.00 x 10<sup>&minus;3</sup> mol</p><p>Since two mol of HCl react with one mol of CaCO<sub>3</sub></p><p>Amount of CaCO<sub>3</sub> that reacts = &frac12; x 4.00 x 10<sup>-3</sup> = 2.00 x&nbsp;10<sup>&minus;3</sup> mol</p><p><em>M</em>(CaCO<sub>3</sub>) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g mol<sup>&minus;1</sup></p><p>Mass of CaCO<sub>3</sub> that reacts with the acid = 100.09 x 2.00 x&nbsp;10<sup>&minus;3</sup> = 0.200 g ( to 3 significant figures).</p><hr class="hidden"><hr class="hidden"><h4>4.The wrong units are used.</h4><p class="MsoNormal">The classic example of where this often occurs is when students use the equation</p><p class="MsoNormal"><em>&Delta;G</em> = <em>&Delta;H </em>&ndash; <em>T&Delta;S</em></p><p class="MsoNormal"><em>&Delta;G</em> and <em>&Delta;H</em> values are usually quoted in kJ mol<sup>&minus;1</sup> but entropy values are normally quoted in J K<sup>&minus;1</sup> mol<sup>&minus;1.</sup> Students constantly need reminding that they must divide the entropy value by 1000 to convert it into kJ K<sup>&minus;1</sup> mol<sup>&minus;1</sup> when using it in the equation. Of course they also need to ensure the temperature is given in Kelvin not <sup>o</sup>C.</p><p class="MsoNormal">Other examples of formulas where units must be consistent are the gas equation <em>pV</em> = <em>nRT</em> and rate equations as the units of the rate constant depend upon the overall order of the reaction.</p><hr class="hidden"><hr class="hidden"><h4>5. Problems with the Henderson&ndash;Hasselbalch equation.</h4><p>This formula causes particular problems although strictly speaking calculations involving buffer solutions should only appear in Paper 3.</p><p>The theory of buffer solutions is covered at Higher Level in the AHL Topic 18 but buffer calculations only appear in two of the options in Paper 3. However even though buffers are not coved at Standard Level in the core, Option D (see <a href="../19184/ph-regulation-of-the-stomach-1.html" title="Options » Option D » pH regulation of the stomach">D.4 pH regulation of the stomach</a>) includes the application of the Henderson&ndash;Hasselbalch equation at both SL and HL whereas in option B (<a href="../19336/proteins-enzymes-ahl--1.html" title="Options » Option B » Proteins &amp; enzymes (AHL) ">B.7 Proteins and enzymes</a>) it is only on the HL syllabus. The use of the Henderson&ndash;Hasselbalch equation in Option D can cause real problems for Standard Level students who have not covered either equilibrium calculations or the definitions of <em>K</em><sub>a</sub>, <em>K</em><sub>b</sub>, p<em>K</em><sub>a</sub> and p<em>K</em><sub>b</sub>.</p><p>The Henderson&ndash;Hasselbalch equation is given in Section 1 of the IB data booklet but it seems unwise for students to use it if they do not fully understand it. It is easy to use wrongly and and I have never taught it to my students as I prefer them to be able to work out all buffer calculations from first principles.</p><hr class="hidden"><h4><span style="color:#FF0000;">Example using first principles</span></h4><p>Calculate the pH of a buffer solution made by adding 20.0 cm<sup>3</sup> of 1.00 mol dm<sup>&minus;3</sup> sodium hydroxide solution to 40.0 cm<sup>3</sup> of 1.00 mol dm<sup>-3</sup> ethanoic acid solution. Section 21 of the IB data booklet gives the p<em>K</em><sub>a</sub> of ethanoic acid = 4.76</p><p>(Standard level students need to understand that <em>K</em><sub>a</sub> is the equilibrium constant (acid dissociation constant) for a weak acid and that p<em>K</em><sub>a</sub> = &minus; log<sub>10</sub><em>K</em><sub>a</sub> (this is rather similar to pH = &minus; log<sub>10</sub>[H<sup>+</sup>(aq)]). The relationship <em>K</em><sub>a</sub> = 10<sup>&minus;p<em>K</em>a</sup> is also helpful as the data booklet only lists p<em>K</em><sub>a</sub> values, not <em>K</em><sub>a</sub> values.)</p><hr class="hidden"><h4><span style="color:#FF0000;">Worked solution:</span></h4><p><strong>Step 1: </strong>Write the equation for the reaction.</p><p>CH<sub>3</sub>COOH(aq) + NaOH(aq) &rarr; CH<sub>3</sub>COONa(aq) + H<sub>2</sub>O(l)</p><p>After the reaction all the sodium hydroxide solution will have been neutralised to form sodium ethanoate and the excess ethanoic acid remains.</p><hr class="hidden"><p><strong>Step 2: </strong>Calculate the amounts of the sodium ethanoate and the excess ethanoic acid in the resulting solution.</p><p>Initial amount of NaOH = 20.0/1000 x 1.00 = 2.00 x 10<sup>&minus;2</sup> mol</p><p>Initial amount of CH<sub>3</sub>COOH = 40.0/1000 x 1.00 = 4.00 x&nbsp;10<sup>&minus;2</sup> mol</p><p>Since all the NaOH reacts to form CH<sub>3</sub>COONa</p><p>Final amount of CH<sub>3</sub>COONa = 2.00 x&nbsp;10<sup>&minus;2</sup> mol</p><p>Final amount of CH<sub>3</sub>COOH = 4.00 x&nbsp;10<sup>&minus;2</sup> &minus; 2.00 x&nbsp;10<sup>&minus;2</sup> = 2.00 x&nbsp;10<sup>&minus;2</sup> mol.</p><hr class="hidden"><p><strong>Step 3:</strong> Calculate the concentrations of the sodium ethanoate and the excess ethanoic acid in the resulting solution.</p><p>Final volume = 20.0 + 40.0 = 60.0 cm<sup>3 </sup> (assuming the volumes are exactly additive).</p><p>[CH<sub>3</sub>COONa(aq)] = 1000 / 60.0 x 2.00 x&nbsp;10<sup>&minus;2</sup> = 0.333&nbsp; mol dm<sup>&minus;3</sup></p><p>[CH<sub>3</sub>COOH(aq)] = 1000 / 60.0 x 2.00 x&nbsp;10<sup>&minus;2</sup> = 0.333&nbsp; mol dm<sup>&minus;3</sup></p><hr class="hidden"><p><strong>Step 4:</strong> Write the equilibrium expression for the dissociation of ethanoic acid.</p><p>CH<sub>3</sub>COOH(aq) ⇌ CH<sub>3</sub>COO<sup>&minus;</sup>(aq) + H<sup>+</sup>(aq)</p><p><em>K</em><sub>a</sub> = [H<sup>+</sup>(aq)] x [CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] / [CH<sub>3</sub>COOH(aq)]</p><hr class="hidden"><p><strong>Step 5: </strong>Substitute the values for the concentration of the acid and the salt into the equilibrium expression to find [H<sup>+</sup>(aq)].</p><p><em>K</em><sub>a </sub>= [H<sup>+</sup>(aq)] x [CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] / [CH<sub>3</sub>COOH(aq)] = [H<sup>+</sup>(aq)] x 0.333 / 0.333 = [H<sup>+</sup>(aq)]</p><p><em>K</em><sub>a </sub>= 10<sup>&minus;4.7</sup> so</p><p>[H<sup>+</sup>(aq)] = 10<sup>&minus;4.76</sup> mol dm<sup>&minus;3</sup></p><hr class="hidden"><p><strong>Step 6: </strong>Convert the hydrogen ion concentration into its pH value</p><p>pH = &minus; log<sub>10</sub>[H<sup>+</sup>(aq)]</p><p>so pH = &minus; log<sub>10</sub>10<sup>&minus;4.76</sup> = 4.76</p><hr class="hidden"><p>Note that this example shows that at the half-equivalence point (i.e. when exactly half of the excess weak acid has been neutralised by the base, so that [acid] = [salt]) the pH = p<em>K</em><sub>a</sub>.</p><hr class="hidden"></div><hr class="hidden"><div class="pinkBg"><h1>Practice questions</h1><p class="MsoNormal"><span style="font-family:Arial">Answer the following questions from first principles (i.e. without using any mathematical formulas). The worked answers can be accessed by clicking on the &lsquo;hidden&rsquo; symbol.</span></p><hr class="hidden"><p class="MsoNormal"><span style="font-family:Arial"></span></p><p class="MsoNormal"><strong>1.</strong><span style="font-family:Arial"> Calculate the concentration of an aqueous solution of sodium nitrate containing 1.70 g of sodium nitrate dissolved in 30.0 cm<sup>3</sup> of the solution. </span></p><section class="tib-hiddenbox"><p>M(NaNO<sub>3</sub>) = 22.99 + 14.01 + (3 x 16.00) = 85.00 g mol<sup>&minus;1</sup></p><p>Amount of NaNO<sub>3</sub> in 30.00 cm<sup>3</sup> = 1.70 &divide; 85.00 = 0.0200 mol</p><p>Amount of NaNO<sub>3</sub> in 1000 cm<sup>3</sup> (1 dm<sup>3</sup>) = 1000 / 30.00 x 0.0200 = &nbsp;0.667 mol</p><p>Concentration of NaNO<sub>3</sub>(aq) = 0.667 mol dm<sup>&minus;3</sup>.</p></section><hr class="hidden"><hr class="hidden"><p><strong>2.</strong>&nbsp;Three separate solutions, 40.0 cm<sup>3</sup> of 2.00 mol dm<sup>&minus;3</sup> HCl(aq), 50.0 cm<sup>3</sup> of 1.60 mol dm<sup>-3</sup> HCl(aq) and 30.0 cm<sup>3</sup> of 1.8 mol dm<sup>&minus;3</sup> HCl(aq) are combined to make one solution. What is the concentration of this new solution?</p><section class="tib-hiddenbox"><p>Amount of HCl in 40.0 cm<sup>3</sup> of 2.00 mol dm<sup>&minus;</sup><sup>3</sup> HCl(aq) = (40.0 / 1000) x 2.00 = 8.00 x 10<sup>&minus;2 </sup>mol.</p><p>Amount of HCl in 50.0 cm<sup>3</sup> of 1.60 mol dm<sup>&minus;</sup><sup>3</sup> HCl(aq) = (50.0 / 1000) x 1.60 = 8.00 x 10<sup>&minus;2 </sup> mol.</p><p>Amount of HCl in 30.0 cm<sup>3</sup> of 1.8 mol dm<sup>&minus;</sup><sup>3</sup> HCl(aq) = (30.0 / 1000) x 1.80 = 5.40 x 10<sup>&minus;2 </sup> mol.</p><p>Total amount of HCl present = (8.00 + 8.00 + 5.40) x 10<sup>&minus;2</sup>&nbsp; = 0.214 mol</p><p>Total volume = 40.0 + 50.0 + 30.0 = 120 cm<sup>3</sup></p><p>Concentration = 0.214 x 1000/120 = 1.78 mol dm<sup>&minus;</sup><sup>3</sup>.</p></section><hr class="hidden"><hr class="hidden"><p><strong>3. </strong>What volume of 0.250 mol dm<sup>&minus;3</sup> sulfuric acid is required to neutralise exactly 15.0 cm<sup>3</sup> of 0.150 mol dm<sup>&minus;3</sup> barium hydroxide solution?</p><section class="tib-hiddenbox"><p>H<sub>2</sub>SO<sub>4</sub>(aq) + Ba(OH)<sub>2</sub> &rarr; BaSO4(aq) + 2H<sub>2</sub>O(l)</p><p>Amount of Ba(OH)<sub>2</sub> in 15.0 cm<sup>3</sup> = (15.0 / 1000) x 0.150 = 2.25 x 10<sup>&minus;3</sup> mol</p><p>1 mol of H<sub>2</sub>SO<sub>4</sub> reacts with 1 mol of Ba(OH)<sub>2</sub></p><p>Amount of H<sub>2</sub>SO<sub>4</sub> required = 2.25 x 10<sup>&minus;3</sup> mol</p><p>Volume of 0.250 mol dm<sup>&minus;3</sup> H<sub>2</sub>SO<sub>4</sub> required = (1000 / 0.250) x 2.25 x 10<sup>&minus;3</sup> = 9.00 cm<sup>3</sup>.</p></section><hr class="hidden"><hr class="hidden"><p><strong>4.&nbsp;</strong>What volume of carbon dioxide measured at STP will be evolved when 2.10 g of sodium hydrogen carbonate, NaHCO<sub>3</sub> is reacted with excess hydrochloric acid?</p><section class="tib-hiddenbox"><p>NaHCO<sub>3</sub>(s) + HCl(aq) &rarr; NaCl(aq) + CO<sub>2</sub>(g) + H<sub>2</sub>O(l)</p><p>M(NaHCO<sub>3</sub>) = 22.99 + 1.01 + 12.01 + (3 x 16.00) = 84.01 g mol<sup>&minus;1</sup></p><p>Amount of NaHCO<sub>3</sub> = 2.10 &divide; 84.01 = 0.0250 mol</p><p>1 mol of NaHCO<sub>3</sub> reacts to produce 1 mol of CO<sub>2</sub></p><p>Amount of CO<sub>2</sub> evolved = 0.0250 mol</p><p>1 mol of any gas occupies 22.7 dm<sup>3</sup> at STP (from Section 2 of the IB data booklet)</p><p>Volume of CO<sub>2</sub> evolved = 0.568 dm<sup>3</sup> = 568 cm<sup>3</sup></p></section><hr class="hidden"><hr class="hidden"><p><strong>5. (a) </strong>What will be the pH of the buffer solution prepared by adding 25.00 cm<sup>3</sup> of 1.25 mol dm<sup>&minus;3</sup> NaOH(aq) to 60.00 cm<sup>3</sup> of 1.20 mol dm<sup>&minus;3</sup> CH<sub>3</sub>COOH(aq). (<em>K</em><sub>a</sub> of CH<sub>3</sub>COOH = 1.74 x 10<sup>&minus;5 </sup>at 298 K)</p><section class="tib-hiddenbox"><p>CH<sub>3</sub>COOH(aq) + NaOH(aq) &rarr; CH<sub>3</sub>COONa(aq) + H<sub>2</sub>O(l)</p><p>Initial amount of NaOH = 25.0/1000 x 1.25 = 3.125 x 10<sup>&minus;2</sup> mol</p><p>Initial amount of CH<sub>3</sub>COOH = 60.0/1000 x 1.20 = 7.200 x&nbsp;10<sup>&minus;2</sup> mol</p><p>Since all the NaOH reacts to form CH<sub>3</sub>COONa</p><p>Final amount of CH<sub>3</sub>COONa = 3.125 x 10<sup>&minus;2</sup> mol</p><p>Final amount of CH<sub>3</sub>COOH = 7.200 x&nbsp;10<sup>&minus;2</sup> &minus; 3.125 x&nbsp;10<sup>&minus;2</sup> = 4.075 x&nbsp;10<sup>&minus;2</sup> mol</p><p>Final volume = 25.0 + 60.0 = 85.0 cm<sup>3 </sup> (assuming the volumes are exactly additive).</p><p>[CH<sub>3</sub>COONa(aq)] = 1000 / 85.0 x 3.125 x 10<sup>&minus;2</sup> = 0.3676&nbsp; mol dm<sup>&minus;3</sup></p><p>[CH<sub>3</sub>COOH(aq)] = 1000 / 85.0 x 4.075 x&nbsp;10<sup>&minus;2</sup> = 0.4794&nbsp; mol dm<sup>&minus;3</sup></p><p><em>K</em><sub>a </sub>= [H<sup>+</sup>(aq)] x [CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] / [CH<sub>3</sub>COOH(aq)] = [H<sup>+</sup>(aq)] x 0.3676 / 0.4794 = 0.767 x [H<sup>+</sup>(aq)]</p><p><em>K</em><sub>a</sub> of CH<sub>3</sub>COOH = 1.74 x 10<sup>&minus;5</sup></p><p>[H<sup>+</sup>(aq)] = 1.74 x 10<sup>&minus;5</sup> / 0.767 = 2.27 x 10<sup>&minus;5</sup></p><p>pH = &minus; log<sub>10</sub>[H<sup>+</sup>(aq)] = 4.64</p></section><p><strong>5.(b) </strong>The same buffer solution can be prepared by dissolving solid sodium ethanoate in a solution of ethanoic acid. What mass of sodium ethanoate would need to be dissolved in 50.0 cm<sup>3</sup> of 1.20 mol dm<sup>&minus;3</sup> CH<sub>3</sub>COOH(aq) to give a buffer solution with a pH of 4.64? Assume that when the salt is dissolved it does not affect the volume of the solution.</p><section class="tib-hiddenbox"><p>CH<sub>3</sub>COOH(aq) ⇌ CH<sub>3</sub>COO<sup>&minus;</sup>(aq) + H<sup>+</sup>(aq)</p><p><em>K</em><sub>a</sub> = [H<sup>+</sup>(aq)] x [CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] / [CH<sub>3</sub>COOH(aq)]</p><p>CH<sub>3</sub>COOH(aq) = 1.20 mol dm<sup>&minus;3</sup> and since the pH must be 4.64, [H<sup>+</sup>(aq)] = 10<sup>&minus;4.64</sup> and the value of <em>K</em><sub>a</sub> is given in <strong>5.(a)</strong>.</p><p><em>K</em><sub>a</sub> = 1.74 x 10<sup>&minus;5</sup> =&nbsp;10<sup>&minus;4.64</sup> x [CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] / 1.20</p><p>[CH<sub>3</sub>COO<sup>&minus;</sup>(aq)] = 1.74 x 10<sup>&minus;5</sup> x 1.20 /&nbsp;10<sup>&minus;4.64</sup> = 0.911 mol dm<sup>&minus;3</sup></p><p><em>M</em>(CH<sub>3</sub>COONa) = (2 x 12.01) + (3 x 1.01) + (2 x 16.00) + 22,99 = 82.04 g mol<sup>&sim;1</sup></p><p>Mass of CH<sub>3</sub>COONa required in 50.0 cm<sup>3</sup> = 0.911 x 82.04 x 50.0 / 1000 = 3.74 g</p></section><hr class="hidden"></div></section></article><section id="media-extras"><div class="page-actions no-print navbar inline hidden-desktop"><div class="navbar-inner"><ul class="nav"><li><a class="presentation" href="#" onclick="return false;"><i class="fa fa-desktop"></i></a></li><li><a class="print-section-blog" href="#" onclick="return false;"><i class="fa fa-print"></i></a></li><li><a class="page-bookmarker" data-ticket="IB Docs (2) Team" data-pid="32586" href="#" onclick="return false;"><i class="fa fa-star"></i></a></li><li><a class="personal-notes" href="#" onclick="return false;"><i class="fa fa-file-text"></i></a></li><li><a class="" data-toggle="modal" href="#modal-feedback" onclick="return false;"><i class="fa fa-envelope-o"></i></a></li><li class="dropdown"><a class="dropdown-toggle" data-toggle="dropdown" data-target="#" href="#"><i class="fa fa-share-alt"></i></a><ul class="dropdown-menu" role="menu"><li><a class="" target="_blank" title="Share on Twitter" href="https://twitter.com/share?text=%22Misuse+of+formulas%22+-+via+%40InThinker+%23chemistry%0A&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32586%2Fmisuse-of-formulas"><i class="fa fa-twitter-square"></i><span>Twitter</span></a></li><li><a class="" target="_blank" title="Share on Facebook" href="https://www.facebook.com/sharer.php?u=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32586%2Fmisuse-of-formulas&t=Misuse+of+formulas"><i class="fa fa-facebook-square"></i><span>Facebook</span></a></li><li><a class="" href="http://www.linkedin.com/shareArticle?mini=true&url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32586%2Fmisuse-of-formulas"><i class="fa fa-linkedin-square"></i><span>LinkedIn</span></a></li><li><a class="" href="https://plus.google.com/share?url=https%3A%2F%2Fwww.thinkib.net%2Fchemistry%2Fpage%2F32586%2Fmisuse-of-formulas"><i class="fa fa-google-plus-square"></i><span>Google +</span></a></li><li><a class="" href="mailto:?subject=Misuse of formulas&body=Students often use mathematical formulas to solve chemical problems in exams and sometimes arrive at the wrong answer. 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