Yep. Chemistry is good ooo, i have enjoyed the tutorial. I really appeciate  the reason why i join this site.

<h2>Titration calculations</h2>
<p>So how exactly can a titration be carried out to determine an unknown concentration? Look at the following steps to help you to understand the process.</p>
<ol>
<li>
<p>A carefully measured volume of the solution with unknown concentration is put into a conical flask.</p>
</li>
<li>
<p>A few drops of a suitable indicator is added to this solution (bromothymol blue and phenolphthalein are common indicators).</p>
</li>
<li>
<p>The conical flask is placed on a white tile or piece of paper (to make colour changes easier to see).</p>
</li>
<li>
<p>A volume of the standard solution (known concentration) is put into a burette (a measuring device) and is slowly added to the solution in the flask, drop by drop.</p>
<img alt="be2fdc082c2cf00ce236416980459bef.png" src="https://data.ulearngo.com/assets/9ed01d7d-4b96-4044-ad37-a07353cc7e8b"/>
<section>
<p>It is better to have the acid in the burette as many <a data-lesson="what-are-acids-and-bases" data-subject="chemistry" data-tutorial="acid-base-redox-reactions" href="https://app.ulearngo.com/chemistry/acid-base-redox-reactions/what-are-acids-and-bases">bases</a> will clog the stopcock and damage the burette. It is much easier to clean a burette that has contained acid only. It is important to make sure the burette is perpendicular to the desk and clamped securely.</p>
<p>When filling the burette the stopcock should be closed and the funnel should be lifted slightly to allow air to escape, otherwise it will bubble through the <a data-lesson="summarizing-properties-of-liquids" data-subject="chemistry" data-tutorial="liquids-and-solids" href="https://app.ulearngo.com/chemistry/liquids-and-solids/summarizing-properties-of-liquids">liquid</a> in the funnel and splash acid. Once the burette is filled a beaker should be placed underneath and a few centimeters of the solution should be drained through the stopcock. This ensures that the entire burette is full. There should be no air bubbles, as a bubble bursting can affect the volume measurement and disrupt the <a data-lesson="measurement-accuracy-precision" data-subject="chemistry" data-tutorial="chemistry-essential-ideas" href="https://app.ulearngo.com/chemistry/chemistry-essential-ideas/measurement-accuracy-precision">accuracy</a> of the experiment.</p>
</section>
</li>
<li>
<p>At some point, adding one more drop will change the colour of the unknown solution to the colour of the end-point of the reaction. Remember the colour changes from the previous lesson on pH scale.</p>
</li>
<li>
<p>Record the volume of standard solution that has been added up to this point.</p>
</li>
<li>
<p>Use the information you have gathered to calculate the exact concentration of the unknown solution. Worked examples are given to walk you through this step.</p>
</li>
<li>
<p>Note that adding more solution once the end-point has been reached will result in a colour change from the end-point colour to that of the acid (if the solution in the conical flask is a base) or of the base (if the solution in the conical flask is an acid).</p>
</li>
</ol>
<p>When you are busy with these calculations, you will need to remember the following:</p>
<p>$\text{1}$ $\text{dm$^{3}$}$ = $\text{1}$ $\text{L}$ = $\text{1 000}$ $\text{mL}$ = $\text{1 000}$ $\text{cm$^{3}$}$, therefore dividing $\text{cm$^{3}$}$ by $\text{1 000}$ will give you an answer in $\text{dm$^{3}$}$.</p>
<p>Some other terms and equations which will be useful to remember are shown below:</p>
<ul>
<li>
<p>concentration of a solution is measured in $\text{mol.dm$^{-3}$}$</p>
</li>
<li>
<p>moles (mol) $=$ concentration ($\text{mol·dm$^{-3}$}$) $\times$ volume ($\text{dm$^{3}$}$)</p>
</li>
<li>\begin{align*} \text{concentration } = \frac{\text{moles}}{\text{volume}} \\ C (\text{mol·dm$^{-3}$}) = \frac{n (\text{mol})}{V (\text{dm$^{3}$})} \end{align*}</li>
<li>
<p>remember to make sure all the units are correct in your calculations</p>
</li>
</ul>
<div class="ns-school-emp">
<h2>Example:</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>Given the equation:</p>
<p>$\text{NaOH}(\text{aq}) + \text{HCl}(\text{aq})$ $\to$ $\text{NaCl}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})$</p>
<p>$\text{25}$ $\text{cm$^{3}$}$ of sodium hydroxide solution was pipetted into a conical flask and titrated with $\text{0.2}$ $\text{mol.dm$^{-3}$}$ hydrochloric acid. Using a suitable indicator, it was found that $\text{15}$ $\text{cm$^{3}$}$ of acid was needed to neutralise the base. Calculate the concentration of the sodium hydroxide.</p>
</div>
<div>
<h3>Step 1: Make sure that the equation is balanced</h3>
<p>There are equal numbers of each type of atom on each <a data-lesson="dissociation-in-water" data-subject="chemistry" data-tutorial="reactions-in-aqueous-solution" href="https://app.ulearngo.com/chemistry/reactions-in-aqueous-solution/dissociation-in-water">side</a> of the equation, so the equation is balanced.</p>
</div>
<div>
<h3>Step 2: Write down all the information you know about the reaction, converting to the correct units</h3>
<p><strong>NaOH:</strong> V = $\text{25}$ $\text{cm$^{3}$}$ $\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$ = $\text{0.025}$ $\text{dm$^{3}$}$</p>
<p><strong>HCl:</strong> V = $\text{15}$ $\text{cm$^{3}$}$ $\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$ = $\text{0.015}$ $\text{dm$^{3}$}$</p>
<p>$\phantom{\rule{25pt}{0ex}}$C = $\text{0.2}$ $\text{mol.dm$^{-3}$}$</p>
</div>
<div>
<h3>Step 3: Calculate the number of moles of HCl that are added</h3>
<p>${\text{C}} = \dfrac{\text{n}}{\text{V}}$</p>
<p>Therefore, n$(\text{HCl})$ = C $\times$ V</p>
<p>n(HCl) = $\text{0.2}$ $\text{mol.dm$^{-3}$}$ $\times$ $\text{0.015}$ $\text{dm$^{3}$}$ = $\text{0.003}$ $\text{mol}$</p>
<p>$\text{0.003}$ $\text{moles}$ of $\text{HCl}$ are required to neutralise the base.</p>
</div>
<div>
<h3>Step 4: Calculate the number of moles of sodium hydroxide in the reaction</h3>
<p>Look at the equation for the reaction: the molar ratio of $\text{HCl}$:$\text{NaOH}$ is $\text{1}$:$\text{1}$.</p>
<p>So for every mole of $\text{HCl}$, there is one mole of $\text{NaOH}$ that is involved in the reaction. Therefore, if $\text{0.003}$ $\text{mol}$ $\text{HCl}$ is required to neutralise the solution, $\text{0.003}$ $\text{mol}$ $\text{NaOH}$ must have been present in the sample of the unknown solution.</p>
</div>
<div>
<h3>Step 5: Calculate the concentration of the sodium hydroxide solution</h3>
<p>${\text{C(NaOH)}} = \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.003}\text{ mol}}{\text{0.025}\text{ dm$^{3}$}} = \text{0.12}\text{ mol.dm$^{-3}$}$</p>
<p>The concentration of the $\text{NaOH}$ solution is $\text{0.12}$ $\text{mol.dm$^{-3}$}$</p>
</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example:</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>$\text{10}$ $\text{g}$ of solid sodium hydroxide is dissolved in $\text{500}$ $\text{cm$^{3}$}$ water. Using titration, it was found that $\text{20}$ $\text{cm$^{3}$}$ of this solution was able to completely neutralise $\text{10}$ $\text{cm$^{3}$}$ of a sulfuric acid solution. Calculate the concentration of the sulfuric acid.</p>
</div>
<div>
<h3>Step 1: Write a balanced equation for the titration reaction</h3>
<p>The reactants are sodium hydroxide ($\text{NaOH}$) and sulfuric acid ($\text{H}_{2}\text{SO}_{4}$). The base has a hydroxide anion ($\text{OH}^{-}$), therefore the products will be a salt and water.</p>
<p>The cation for the salt ($\text{Na}^{+}$) will come from the base. The anion for the salt ($\text{SO}_{4}^{2-}$) will come from the acid. There must be $\text{2}$ $\text{Na}^{+}$ cations for every one $\text{SO}_{4}^{2-}$ and the salt will be $\text{Na}_{2}\text{SO}_{4}$.</p>
<p>$\text{H}_{2}\text{SO}_{4}(\text{aq}) + \text{NaOH}(\text{aq})$ $\to$ $\text{Na}_{2}\text{SO}_{4}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})$</p>
<p>To balance the equation we need to multipy the number of sodium hydroxide molecules and the water molecules by two.</p>
<p>$\text{H}_{2}\text{SO}_{4}(\text{aq}) + 2\text{NaOH}(\text{aq})$ $\to$ $\text{Na}_{2}\text{SO}_{4}(\text{aq}) + 2\text{H}_{2}\text{O}(\text{l})$</p>
</div>
<div>
<h3>Step 2: Calculate the concentration of the sodium hydroxide solution</h3>
<p>The total volume that the $\text{10}$ $\text{g}$ was dissolved in must be used to calculate the concentration.</p>
<p>V = $\text{500}$ $\text{cm$^{3}$}$ $\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$ = $\text{0.5}$ $\text{dm$^{3}$}$</p>
<p>M(NaOH) = $\text{23.0}$ + $\text{16.0}$ + $\text{1.01}$ = $\text{40.01}$ $\text{g.mol$^{-1}$}$</p>
<p>n(NaOH) = $\dfrac{\text{m}}{\text{M}} = \dfrac{\text{10}\text{ g}}{\text{40.01}\text{ g.mol$^{-1}$}} = \text{0.25}\text{ mol}$</p>
<p>C(NaOH) = $\dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.25}\text{ mol}}{\text{0.5}\text{ dm$^{3}$}} = \text{0.50}\text{ mol.dm$^{-3}$}$</p>
</div>
<div>
<h3>Step 3: Calculate the number of moles of sodium hydroxide that were used in the neutralisation reaction</h3>
<p>Remember that only $\text{20}$ $\text{cm$^{3}$}$ of the sodium hydroxide solution is used:</p>
<p>V = $\text{20}$ $\text{cm$^{3}$}$ $\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$ = $\text{0.02}$ $\text{dm$^{3}$}$</p>
<p>$\text{C} = \dfrac{\text{n}}{\text{V}}$, therefore $\text{n} = {\text{C}} \times {\text{V}}$</p>
<p>$\text{n} = \text{0.50}\text{ mol.dm$^{-3}$} \times \text{0.02}\text{ dm$^{3}$} = \text{0.01}\text{ mol}$</p>
</div>
<div>
<h3>Step 4: Calculate the number of moles of sulfuric acid that were neutralised</h3>
<p>According to the balanced chemical equation, <a data-lesson="the-mole-2" data-subject="chemistry" data-tutorial="quantitative-aspects-chemical-change" href="https://app.ulearngo.com/chemistry/quantitative-aspects-chemical-change/the-mole-2">the mole</a> ratio of $\text{NaOH}$ to $\text{H}_{2}\text{SO}_{4}$ is $\text{2}$:$\text{1}$. There are $\text{2}$ moles of $\text{NaOH}$ for every $\text{1}$ mole of $\text{H}_{2}\text{SO}_{4}$.</p>
<p>n($\text{H}_{2}\text{SO}_{4}$ = $\dfrac{\text{0.01}\text{ mol}}{\text{2}}$ = $\text{0.005}$ $\text{mol}$.</p>
</div>
<div>
<h3>Step 5: Calculate the concentration of the sulfuric acid solution</h3>
<p>Remember that $\text{10}$ $\text{cm$^{3}$}$ of the sulfuric acid solution is neutralised.</p>
<p>V($\text{H}_{2}\text{SO}_{4}$) = $\text{10}$ $\text{cm$^{3}$}$ $\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$ = $\text{0.01}$ $\text{dm$^{3}$}$</p>
<p>$\text{C} = \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.005}\text{ mol}}{\text{0.01}\text{ dm$^{3}$}} = \text{0.5}\text{ mol.dm$^{-3}$}$</p>
</div>
</div>
</div>

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Titration Calculations

Chemistry generally explores the substances of which matter is composed, the investigation of their properties and reactions, and the use of such reactions to form new substances. 

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Titration Calculations

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