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\author{NescafeAbusive32 (nescafeabusive32)}
\pdfinfo{
  /Title (gr-12-chemical-systems-and-equilibrium.pdf)
  /Creator (Cheatography)
  /Author (NescafeAbusive32 (nescafeabusive32))
  /Subject (Gr. 12 Chemical Systems and Equilibrium Cheat Sheet)
}

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        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Gr. 12 Chemical Systems and Equilibrium Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{NescafeAbusive32 (nescafeabusive32)} via \textcolor{DarkBackground}{\uline{cheatography.com/53385/cs/14454/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}NescafeAbusive32 (nescafeabusive32) \\
  \uline{cheatography.com/nescafeabusive32} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 25th January, 2018.\\
   Updated 27th January, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
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\begin{document}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{p{0.4977 cm} p{0.4977 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Introduction}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Equilibrium}} = the point in a chemical reaction where the reactants and the products are formed and broken {\bf{at the same rate}}} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Dynamic equilibrium}} = a balance between the forward and backward rates that are {\bf{occurring simultaneously}}} \tn 
% Row Count 6 (+ 3)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Equilibrium law}} = {\bf{mathematical description}} of a chemical system {\bf{at equilibrium}}} \tn 
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% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Equilibrium constant}} (K or K`eq`) = the numerical value {\bf{defining the equilibrium law}} for a system {\bf{at a given temperature}} (changes with temperature)} \tn 
% Row Count 12 (+ 4)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Heterogeneous equilibrium}} = products and reactants are in {\bf{at least 2 different states}}; {\bf{pure solids/liquids}} are {\bf{not}} included in K`eq` formula} \tn 
% Row Count 16 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.38896 cm} x{2.58804 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Equilibrium Constant (K`eq`)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\bf{K`eq` formula:}}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{If {\bf{a{[}A{]} + b{[}B{]} ⇌ c{[}C{]} + d{[}D{]}}}; then} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}{\bf{K`eq` = ({[}C{]}\textasciicircum{}c\textasciicircum{}{[}D{]}\textasciicircum{}d\textasciicircum{}) ÷ ({[}A{]}\textasciicircum{}a\textasciicircum{}{[}B{]}\textasciicircum{}b\textasciicircum{})}}} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\emph{Magnitude of K`eq`:}} states whether the {\bf{equilibrium position favours products/reactants}}} \tn 
% Row Count 5 (+ 2)
% Row 4
\SetRowColor{LightBackground}
If K = 1 & {[}products{]} = {[}reactants{]} \tn 
% Row Count 7 (+ 2)
% Row 5
\SetRowColor{white}
If K\{\{fa-angle-double-right\}\}1 & {[}products{]} \{\{fa-angle-double-right\}\} {[}reactants{]} \tn 
% Row Count 10 (+ 3)
% Row 6
\SetRowColor{LightBackground}
If K\{\{fa-angle-double-left\}\}1 & {[}products{]} \{\{fa-angle-double-left\}\} {[}reactants{]} \tn 
% Row Count 13 (+ 3)
% Row 7
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\bf{K`forward` vs K`backward`}}} \tn 
% Row Count 14 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}If {\bf{a{[}A{]} ⇌ b{[}B{]}}}, then} \tn 
% Row Count 15 (+ 1)
% Row 9
\SetRowColor{white}
{\bf{K`forward` = ({[}B{]}\textasciicircum{}b\textasciicircum{}) ÷ ({[}A{]}\textasciicircum{}a\textasciicircum{})}} & {\bf{K`backward` = ({[}A{]}\textasciicircum{}a\textasciicircum{}) ÷ ({[}B{]}\textasciicircum{}b\textasciicircum{})}} \tn 
% Row Count 17 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}So {\bf{∴ K`forward` = \textasciicircum{}1\textasciicircum{}/`Kbackward`}} @ equilibrium} \tn 
% Row Count 19 (+ 2)
% Row 11
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\emph{Purpose of K`eq`:}} to {\bf{determine equilibrium concentration}} of chemical entities {\bf{given initial conditions}} (I.C.E. table)} \tn 
% Row Count 22 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Reaction Quotient (Q)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Helps to {\bf{determine the position of the equilibrium}} of a system using the {\bf{rate law}} for the system and comparing it with the K`eq`} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
If Q \textless{} K`eq` ~ ~ ~ ~ ~ ~ ~ ~ & {[}products{]} \textless{} {[}reactants{]}\{\{nl\}\}Reaction has not reached ⇌ yet; reaction needs to {\bf{shift right}} \tn 
% Row Count 8 (+ 5)
% Row 2
\SetRowColor{LightBackground}
If Q \textgreater{} K`eq` & {[}products{]} \textgreater{} {[}reactants{]}\{\{nl\}\}Reaction has not reached ⇌ yet; reaction needs to {\bf{shift left}} \tn 
% Row Count 13 (+ 5)
% Row 3
\SetRowColor{white}
If Q = K`eq` & {[}products{]} = {[}reactants{]}\{\{nl\}\}Reaction has not reached equilibrium yet; {\bf{no shift}}  will occur \tn 
% Row Count 18 (+ 5)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Variables Affecting Chemical Equilibria}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Le Châtelier's Principle}}: When a chemical system {\bf{at equilibrium}} is {\bf{disturbed}} by a {\bf{change in property}}, the system responds in a way that {\bf{opposes the change}}} \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}{\bf{Concentration/Temperature}}} \tn 
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% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}\{\{fa-long-arrow-up\}\}{[}conc{]}/T = shift to {\bf{consume}}} \tn 
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% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{fa-long-arrow-down\}\}{[}conc{]}/T = shift to {\bf{replace}}} \tn 
% Row Count 9 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{If you {\bf{add more}} reactant/heat to a system, the system will {\bf{consume it}} to make more product, and {\bf{vice versa}}} \tn 
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% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{If you {\bf{remove}} reactant/heat from a system, the system will {\bf{replace it}} from the existing product, and {\bf{vice versa}}} \tn 
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% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}{\bf{Volume/Pressure (gases only)}}} \tn 
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% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}\{\{fa-long-arrow-up\}\}V =\{\{fa-long-arrow-down\}\}P = shift toward side with {\bf{more}} gas entities (i.e. more mol of gas) (more space for particles)} \tn 
% Row Count 19 (+ 3)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{fa-long-arrow-down\}\}V =\{\{fa-long-arrow-up\}\}P = shift toward side with {\bf{fewer}} gas entities (i.e. less mol of gas) (less space for particles)} \tn 
% Row Count 22 (+ 3)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}{\bf{Catalysts/Inert (noble) gases}} \{\{fa-long-arrow-right\}\} No effect} \tn 
% Row Count 24 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Equilibria of Slightly Soluble Compounds}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Molar solubility}}: The {\bf{amount of solute}} (in mol) that can be {\bf{dissolved in 1 L}} of solvent at a {\bf{certain temperature}}} \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\bf{T α solubility:}} as temperature increases, so does molar solubility} \tn 
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% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{All compounds have {\bf{some solubility}}, even if they are considered "insoluble" (really {\bf{very low molar solubility}})} \tn 
% Row Count 8 (+ 3)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\bf{Very soluble}} compounds ({\bf{high}} molar solubility) = {\bf{no ⇌}} ({\bf{complete disassociation}} into ions)} \tn 
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% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\bf{Slightly soluble}} compounds ({\bf{low}} molar solubility) = {\bf{has ⇌}} ({\bf{incomplete/partial disassociation}} into ions)} \tn 
% Row Count 14 (+ 3)
% Row 5
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{As a compound is placed in a solvent, {\bf{some part of it will dissolve}}, but at the same time, the {\bf{reverse reaction starts}}} \tn 
% Row Count 17 (+ 3)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Eventually the two reactions reach equilibrium, creating a {\bf{saturated solution}} (at this point, {[}conc{]} of ions remains {\bf{constant}}) \{\{fa-long-arrow-right\}\} solubility equilibrium} \tn 
% Row Count 21 (+ 4)
% Row 7
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\bf{Equilibrium formula for solubility:}}} \tn 
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% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{If {\bf{x{[}A`a`B`b`{]} `(s)` ⇌ a{[}A\textasciicircum{}b+\textasciicircum{}{]} `(aq)` + b{[}B\textasciicircum{}a-\textasciicircum{}{]} `(aq)`}}, then} \tn 
% Row Count 24 (+ 2)
% Row 9
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}{\bf{K`sp` = {[}A\textasciicircum{}b+\textasciicircum{}{]}\textasciicircum{}a\textasciicircum{}{[}B\textasciicircum{}a-\textasciicircum{}{]}\textasciicircum{}b\textasciicircum{}}}} \tn 
% Row Count 25 (+ 1)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{K`sp` (Solubility product constant):}} the product of the {[}conc{]} of ions in a saturated solution} \tn 
% Row Count 27 (+ 2)
% Row 11
\SetRowColor{white}
If {\bf{K`sp`\{\{fa-angle-double-left\}\}1}} & Amount (mol) of aqueous ions{\bf{\{\{fa-angle-double-left\}\}}}amount (mol) of solid substance \tn 
% Row Count 32 (+ 5)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
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\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Equilibria of Slightly Soluble Compounds (cont)}}  \tn
% Row 12
\SetRowColor{LightBackground}
If {\bf{K`sp`\{\{fa-angle-double-right\}\}1}} & Amount (mol) of aqueous ions{\bf{\{\{fa-angle-double-right\}\}}}amount (mol) of solid substance \tn 
% Row Count 5 (+ 5)
% Row 13
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bb\}\}Molar solubility and the K`sp` {\bf{describe the solubility of a substance}} in different ways, meaning you can use one to solve for the other (using an ICE table)} \tn 
% Row Count 9 (+ 4)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{bb\}\}{\bf{Using Q and K`sp` to predict precipitate formation}}} \tn 
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% Row 15
\SetRowColor{white}
If Q \textless{} K`sp` ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ & Not enough ions in solution (unsaturated); {\bf{no precipitate}}; reaction will shift to the {\bf{right}} \tn 
% Row Count 17 (+ 6)
% Row 16
\SetRowColor{LightBackground}
If Q \textgreater{} K`sp` & Lots of ions in solution (saturated); {\bf{precipitate can form}}; reaction will shift to the {\bf{left}} \tn 
% Row Count 22 (+ 5)
% Row 17
\SetRowColor{white}
If Q = K`sp` & System is at ⇌; {\bf{no precipitate}} (saturated solution) \tn 
% Row Count 25 (+ 3)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.53827 cm} x{2.43873 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{pH and pOH}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{pH/pOH:}} The measure of {\bf{acidity/alkalinity}} of a solution} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{pH:}} The measure of {\bf{{[}H\textasciicircum{}+\textasciicircum{}{]}}} in a solution} \tn 
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% Row 2
\SetRowColor{LightBackground}
{\bf{pH = -log{[}H\textasciicircum{}+\textasciicircum{}{]}\{\{nl\}\}= -log{[}H`3`O\textasciicircum{}+\textasciicircum{}{]}}} & {\bf{{[}H`3`O\textasciicircum{}+\textasciicircum{}{]} = {[}H\textasciicircum{}+\textasciicircum{}{]}\{\{nl\}\}= 10\textasciicircum{}-pH\textasciicircum{}}} \tn 
% Row Count 6 (+ 3)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{pOH:}} The measure of {\bf{{[}OH¯{]}}} in a solution} \tn 
% Row Count 7 (+ 1)
% Row 4
\SetRowColor{LightBackground}
{\bf{pH = -log{[}OH¯{]}}} & {\bf{{[}OH¯{]} = 10\textasciicircum{}-pOH\textasciicircum{}}} \tn 
% Row Count 9 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{The pH and pOH values are related to the exponent of K`w` (14):} \tn 
% Row Count 11 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}{\bf{pH + pOH = 14}}} \tn 
% Row Count 12 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Acids and Bases}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Arrhenius acid/base}} = solution that {\bf{ionizes}} into {\bf{H\textasciicircum{}+\textasciicircum{}}} (acid)/{\bf{OH¯}} (base) {\bf{ions}}} \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Bronsted-Lowry acid/base}} = solution that {\bf{donates}} (acid)/{\bf{receives}} (base) {\bf{H\textasciicircum{}+\textasciicircum{} ions}}} \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Strong acid/base:}} solution that {\bf{completely ionizes}} (acid)/{\bf{disassociates}} (base) into ions} \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Weak acid/base:}} solution that {\bf{partially ionizes}} (acid)/{\bf{disassociates}} (base) into ions} \tn 
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% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Monoprotic acids:}} acids that donate {\bf{one H\textasciicircum{}+\textasciicircum{} ion}}} \tn 
% Row Count 10 (+ 2)
% Row 5
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Polyprotic acids:}} acids that donate {\bf{more than one H\textasciicircum{}+\textasciicircum{} ion}}\{\{nl\}\}({\emph{diprotic}} = 2 H\textasciicircum{}+\textasciicircum{}, {\emph{triprotic}} = 3 H\textasciicircum{}+\textasciicircum{}, etc.)} \tn 
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% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{{\emph{Amphiprotic substance:}} substance that can behave like an acid or as a base (i.e. can {\bf{donate and receive}} H\textasciicircum{}+\textasciicircum{} ions)} \tn 
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% Row 7
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\bf{Neutralization reactions}}} \tn 
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% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{With {\bf{strong}} acids/bases:} \tn 
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% Row 9
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\} acid + base \{\{fa-long-arrow-right\}\} salt + water} \tn 
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% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{*{\bf{Complete}} ionization, so {\bf{no equilibrium analysis}}*} \tn 
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% Row 11
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{With {\bf{weak}} acids/bases:} \tn 
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% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\} acid + base ⇌ {\bf{conjugate base}} + {\bf{conjugate acid}}} \tn 
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% Row 13
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{*{\bf{Partial}} ionization, so {\bf{equilibrium has to be analyzed}}*} \tn 
% Row Count 27 (+ 2)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}{\bf{Acid/Base Constants (K`a` and K`b`)}}} \tn 
% Row Count 28 (+ 1)
% Row 15
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{bt\}\}K`a`/K`b` indicate the {\bf{strength}} of an acid/base} \tn 
% Row Count 30 (+ 2)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Acids and Bases (cont)}}  \tn
% Row 16
\SetRowColor{LightBackground}
If K`a`/K`b`\{\{fa-angle-double-right\}\}1 ~ ~ ~ ~ ~ ~ ~ ~ & Strong acid/base ({\bf{complete}} \seqsplit{ionization/disassociation)} \tn 
% Row Count 5 (+ 5)
% Row 17
\SetRowColor{white}
If K`a`/K`b`\{\{fa-angle-double-left\}\}1 & Weak acid/base ({\bf{partial}} \seqsplit{ionization/disassociation)} \tn 
% Row Count 8 (+ 3)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{If one K is known, the other can be determined using K`w` through the formula:} \tn 
% Row Count 10 (+ 2)
% Row 19
\SetRowColor{white}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}{\bf{K`w` = K`a`}} (of acid) {\bf{⋅ K`b`}} (of conjugate base)} \tn 
% Row Count 12 (+ 2)
% Row 20
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{\{\{ac\}\}{\bf{K`w` = K`b`}} (of base) {\bf{⋅ K`a`}} (of conjugate acid)} \tn 
% Row Count 14 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Autoionization of Water and Water Constant (K`w`)}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Water can {\bf{dissociate}} into ions {\bf{on its own:}}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{H`2`O`(l)` ⇌ H\textasciicircum{}+\textasciicircum{}`(aq)` + OH¯`(aq)`}}} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{But the H\textasciicircum{}+\textasciicircum{} ion can also {\bf{attack other H`2`O molecules:}}} \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{H`2`O`(l)` + H\textasciicircum{}+\textasciicircum{}`(aq)` ⇌ H`3`O\textasciicircum{}+\textasciicircum{}`(aq)`}}\textasciicircum{}1\textasciicircum{}} \tn 
% Row Count 6 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Adding both reactions together:} \tn 
% Row Count 7 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{2 H`2`O`(l)` ⇌ H`3`O\textasciicircum{}+\textasciicircum{}`(aq)` + OH¯`(aq)`}}} \tn 
% Row Count 9 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{All equilibria have a constant (K) value, therefore:} \tn 
% Row Count 11 (+ 2)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{K`w`}} = {\bf{{[}H`3`O\textasciicircum{}+\textasciicircum{}{]}{[}OH¯{]}}}\{\{nl\}\}(H`2`O`(l)` not included because it is {\bf{not `(aq)`}})} \tn 
% Row Count 13 (+ 2)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{bt\}\}Since water is {\bf{neutral}} (pH = 7):} \tn 
% Row Count 14 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{{[}H\textasciicircum{}+\textasciicircum{}{]} = 1.0⋅10\textasciicircum{}-7\textasciicircum{}}} \{\{fa-long-arrow-right\}\} {\bf{{[}H`3`O\textasciicircum{}+\textasciicircum{}{]} = 1.0⋅10\textasciicircum{}-7\textasciicircum{}}}} \tn 
% Row Count 16 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{pH + pOH = 14}} \{\{fa-long-arrow-right\}\} {\bf{pOH = 7}} \{\{fa-long-arrow-right\}\}\{\{nl\}\} {\bf{{[}OH¯{]} = 1.0⋅10\textasciicircum{}-7\textasciicircum{}}}} \tn 
% Row Count 19 (+ 3)
% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{If {\bf{{[}H`3`O\textasciicircum{}+\textasciicircum{}{]} = 1.0⋅10\textasciicircum{}-7\textasciicircum{}}}, and {\bf{{[}OH¯{]} = 1.0⋅10\textasciicircum{}-7\textasciicircum{}}}, then:} \tn 
% Row Count 21 (+ 2)
% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\{\{ac\}\}{\bf{K`w`}} = {\bf{(1.0⋅10\textasciicircum{}-7\textasciicircum{})(1.0⋅10\textasciicircum{}-7\textasciicircum{})\{\{nl\}\}= 1.0⋅10\textasciicircum{}-14\textasciicircum{}}}*} \tn 
% Row Count 23 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{\textasciicircum{}{[}1{]}\textasciicircum{}H`3`O\textasciicircum{}+\textasciicircum{}`(aq)`: {\bf{Hydronium ion}} \newline  \newline * Value of K`w` is {\bf{always}} 1.0⋅10\textasciicircum{}-14\textasciicircum{}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}