How can you use ICE tables to solve multiple coupled equilibria? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-08-18T23:20:31Z https://chemistry.stackexchange.com/feeds/question/111041 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://chemistry.stackexchange.com/q/111041 6 How can you use ICE tables to solve multiple coupled equilibria? Karsten Theis https://chemistry.stackexchange.com/users/72973 2019-03-15T18:58:03Z 2019-03-18T22:39:49Z <p>If I have a problem involving multiple coupled equilibrium reactions, such as</p> <blockquote> <p>Calcium fluoride, <span class="math-container">$\ce{CaF2}$</span>, has a molar solubility of <span class="math-container">$\pu{2.1e−4 mol L−1}$</span> at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p<span class="math-container">$K_{\mathrm{a}}$</span> of <span class="math-container">$\ce{HF}$</span> is 3.17.</p> </blockquote> <p>The relevant reactions are:</p> <p><span class="math-container">$$\ce{CaF2(s) &lt;=&gt; Ca^2+(aq) + 2 F-(aq)}$$</span> and <span class="math-container">$$\ce{HF(aq) &lt;=&gt; H+(aq) + F-(aq)}$$</span></p> <p>They are coupled because fluoride occurs in both of them.</p> <p>Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?</p> <p>For example, I could try to set up one ICE table for each reaction (the column for <span class="math-container">$\ce{H+}$</span> is strange because in the problem, the pH is set to a value through an unspecified mechanism):</p> <p><span class="math-container">$$\begin{array}{|c|c|c|} \hline &amp;[\ce{Ca^2+}] &amp; [\ce{F-}] \\ \hline I &amp; \pu{2.1e−4} &amp; \pu{4.2e−4} \\ \hline C &amp; +x &amp; +2x \\ \hline E &amp; \pu{2.1e−4}+x &amp; \pu{4.2e−4}+2x \\ \hline \end{array}$$</span></p> <p>and</p> <p><span class="math-container">$$\begin{array}{|c|c|c|} \hline &amp;[\ce{HF}] &amp; [\ce{H+}] &amp; [\ce{F-}] \\ \hline I &amp; 0 &amp; \text{N/A} &amp; \pu{4.2e−4} \\ \hline C &amp; +x &amp;\text{N/A} &amp; -x \\ \hline E &amp; +x &amp; 10^{-3.00} &amp; \pu{4.2e−4} - x\\ \hline \end{array}$$</span></p> <p>However, how do the fluoride concentrations "talk to each other" in the two tables? Is the <span class="math-container">$x$</span> in one table the same as the <span class="math-container">$x$</span> in the other table?</p> https://chemistry.stackexchange.com/questions/111041/-/111042#111042 8 Answer by Karsten Theis for How can you use ICE tables to solve multiple coupled equilibria? Karsten Theis https://chemistry.stackexchange.com/users/72973 2019-03-15T18:58:03Z 2019-03-15T21:32:16Z <p>The way to use ICE tables in this case would be to combine the ICE tables, and use "<span class="math-container">$x$</span>" for the changes due to one reaction and "<span class="math-container">$y$</span>" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, <em>i.e.</em> <span class="math-container">$x$</span> and <span class="math-container">$y$</span>), so you need two pieces of information to solve it, in this case the two equilibrium constants.</p> <p>Here is the combined ICE table:</p> <p><span class="math-container">$$\begin{array}{|c|c|c|c|c|} \hline &amp;[\ce{Ca^2+}] &amp; [\ce{F-}] &amp; [\ce{H+}]&amp;[\ce{HF}] \\ \hline I &amp; \pu{2.1e−4} &amp; \pu{4.2e−4} &amp; \text{N/A} &amp; 0 \\ \hline C &amp; +x &amp; +2x-y &amp; \text{N/A} &amp; +y \\ \hline E &amp; \pu{2.1e−4}+x &amp; \pu{4.2e−4}+2x-y &amp; 10^{-3.00} &amp; +y \\ \hline \end{array}$$</span></p> <p>Now, you can use the equilibrium constants for the two reactions to solve for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, giving you the equilibrium concentrations.</p> <p>What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.</p> <p>Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using <span class="math-container">$p$</span> and <span class="math-container">$q$</span> because they will be different from <span class="math-container">$x$</span> and <span class="math-container">$y$</span>):</p> <p><span class="math-container">$$\begin{array}{|c|c|c|c|c|} \hline &amp;[\ce{Ca^2+}] &amp; [\ce{F-}] &amp; [\ce{H+}]&amp;[\ce{HF}] \\ \hline I &amp; 0 &amp; 0 &amp; \text{N/A} &amp; 0 \\ \hline C &amp; +p &amp; +2p-q &amp; \text{N/A} &amp; +q \\ \hline E &amp; +p &amp; +2p-q &amp; 10^{-3.00} &amp; +q \\ \hline \end{array}$$</span></p> https://chemistry.stackexchange.com/questions/111041/-/111192#111192 1 Answer by Zhe for How can you use ICE tables to solve multiple coupled equilibria? Zhe https://chemistry.stackexchange.com/users/35555 2019-03-18T22:13:49Z 2019-03-18T22:39:49Z <p>Alternatively, given the cross promotion of a <a href="https://chemistry.stackexchange.com/questions/110670/can-we-combine-two-reactions-and-then-calculate-the-equilibrium-concentrations">related question</a>, we can consider doing away with ICE tables all together.</p> <p><span class="math-container">$$K_{\mathrm{sp}} = \ce{[Ca^{2+}][F-]^{2}}=\pu{3.7e−11}$$</span> <span class="math-container">$$K_{a} = \frac{\ce{[H+][F-]}}{\ce{[HF]}}=10^{-3.17}$$</span></p> <p>At this point, the unknowns are <span class="math-container">$\ce{[Ca^{2+}]}$</span>, <span class="math-container">$\ce{[F-]}$</span>, <span class="math-container">$\ce{[HF]}$</span>. <span class="math-container">$\ce{[H+]}$</span> is known from the pH and it's <span class="math-container">$10^{-3}$</span>.</p> <p>We have two equations and 3 unknowns. The final piece that's missing is that because the source of all fluorine and calcium in this system is from calcium fluoride, we may impose the stoichiometry of the solid on the species present:</p> <p><span class="math-container">$$2\ce{[Ca^{2+}]} = \ce{[F-]} + \ce{[HF]}$$</span></p> <p>Three equations and three unknowns. I cheated and used WolframAlpha:</p> <p><span class="math-container">$$\ce{[Ca^{2+}]} \approx 3.8\times 10^{-4}$$</span> This represents an approximately 1.8 fold increase in solubility.</p> <p>(I sanity checked the calcium ion concentration for a neutral solution using this method and got back the molar solubility.)</p>