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% Document Info
\author{nanana00}
\pdfinfo{
  /Title (chemistry-acids-and-bases.pdf)
  /Creator (Cheatography)
  /Author (nanana00)
  /Subject (Chemistry - Acids \& Bases Cheat Sheet)
}

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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Chemistry - Acids \& Bases Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{nanana00} via \textcolor{DarkBackground}{\uline{cheatography.com/145954/cs/31491/}}}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}nanana00 \\
  \uline{cheatography.com/nanana00} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 20th April, 2022.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{x{2.8 cm} x{5.2 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{ACID}}  \tn
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\SetRowColor{LightBackground}
PROTON DONOR & H+ \tn 
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\SetRowColor{white}
Arrhenius Concept & increases H+ in water (ex. H3O+) \tn 
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\SetRowColor{LightBackground}
\seqsplit{Bronsted-Lowry} Concept & can donate a proton (H+) \tn 
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\SetRowColor{white}
Lewis Concept & electron pair acceptor (covalent bond created) \tn 
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{\emph{strong acids}} & dissociate fully (100\% ionized) \tn 
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{\emph{weak acids}} & dissociate partially (\textless{}100\% ionized) \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{tart/sour taste \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{x{3.04 cm} x{4.96 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{ACID/BASE REACTIONS}}  \tn
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\SetRowColor{LightBackground}
can be ions or substances & not limited to (aq) solutions \tn 
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\mymulticolumn{2}{x{8.4cm}}{some react as acid or base depending on other reactant} \tn 
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\SetRowColor{LightBackground}
salt & ionic compound, product of acid base reaction, does {\bf{NOT}} contain OH-, not metal oxide \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{ex. HCl (aq) + KOH (aq) -{}-{}-{}-\textgreater{} KCl (aq) + H2O \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{AUTOIONIZATION OF WATER}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{small \% of water undergoes ionization to produce ions \{\{ac\}\}} \tn 
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\mymulticolumn{1}{x{8.4cm}}{Kw = 1 × 10\textasciicircum{}-14 \{\{ac\}\}} \tn 
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% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{ex. H2O (l) + H2O (l) -{}-\textgreater{} H3O+ (aq) + OH- (aq) \{\{ac\}\}} \tn 
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\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{in pure water.... {[}H3O+{]} × {[}OH-{]} = 10\textasciicircum{}-14   and   {[}H3O+{]} = {[}OH-{]} = 10\textasciicircum{}-7 \{\{ac\}\}} \tn 
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\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{{\emph{the product between the molar concentrations of {\bf{hydronium ion}} and {\bf{hydroxide ion}} is a constant}} \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{8.4cm}{x{4.08 cm} x{3.92 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{pH SCALE}}  \tn
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ACID & BASE \tn 
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pH = -log{[}H3O+{]}  OR  {[}H3O+{]} = 10\textasciicircum{}-pH & pOH = -log{[}OH-{]}  OR   {[}OH-{]} =10\textasciicircum{}-pH \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{pH = -log{[}H+{]} \newline -pH = log{[}H3O+{]} \newline 10\textasciicircum{}    \textasciicircum{}10 \newline {[}H+{]} = 10\textasciicircum{}-pH}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\begin{tabularx}{8.4cm}{x{4 cm} x{4 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{ACID IONIIZATION CONSTANT, Ka}}  \tn
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\mymulticolumn{2}{x{8.4cm}}{} \tn 
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{\emph{WEAK ACID IONIZATION}} & equilibria of weak acid \tn 
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HA (aq) + H2O⇄H3O\textasciicircum{}+\textasciicircum{}(aq) + A- & {\bf{Ka = {[}H+{]}{[}A-{]}/{[}HA{]}}} \tn 
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\mymulticolumn{2}{x{8.4cm}}{pKa = -logKa} \tn 
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pH = -log {[}H3O+{]} & {[}H3O+{]} = 10\textasciicircum{}-pH\textasciicircum{} \tn 
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pOH = -log{[}OH-{]} & {[}OH-{]} = 10\textasciicircum{}-pH\textasciicircum{} \tn 
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\mymulticolumn{2}{x{8.4cm}}{p = -log} \tn 
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Degree of Ionization, α & neutral molecule splits into charged ions when exposed in a solution \tn 
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α of weak acid/base in water = fraction of total concentration that has formed ions & ex. HA(aq) + H2O⇄H3O\textasciicircum{}+\textasciicircum{}(aq) + A- (aq) \tn 
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\% ionization ({\emph{weak acid}}) = & {[}A-{]} eq/{[}HA{]} orginally × 100\% \tn 
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\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{the larger the Ka, the stronger the acid \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Ka Table}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/nanana00_1649686495_speciationmodels.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{8.4cm}{x{2.8 cm} x{5.2 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{BASE}}  \tn
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\SetRowColor{LightBackground}
PROTON RECEPTOR & OH- \tn 
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% Row 1
\SetRowColor{white}
Arrhenius Concept & increase {[}OH-{]} in water \tn 
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\SetRowColor{LightBackground}
\seqsplit{Bronsted-Lowry} Concept & can accept a proton (H+) \tn 
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\SetRowColor{white}
Lewis Concept & electron pair donor (covalent bond created) \tn 
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\SetRowColor{LightBackground}
{\emph{strong bases}} & dissociate fully (100\% ionized) \tn 
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\SetRowColor{white}
{\emph{weak bases}} & dissociate partially (\textgreater{}100\%) \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{bitter taste, slippery feeling \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{BRONSTED ACID/BASE STRENGHT}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{8.4cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/nanana00_1649017710_chepic.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{acid 1 + base 2 \textless{}-{}-\textgreater{} (conjugate) base 1 + (conjugate) acid 2 \{\{ac\}\} \newline rxn direction favors weaker acid and base \{\{ac\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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\begin{tabularx}{8.4cm}{x{4.08 cm} x{3.92 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{AUTOIONIZATION OF WATER PROBLEMS}}  \tn
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\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{1. {\bf{hydrogen (hydronium) ion concentration when {[}OH-{]} = 2× 10\textasciicircum{}-3\textasciicircum{}M}} \{\{ac\}\}} \tn 
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 & Kw = {[}H+{]}{[}OH-{]} = 1 × 10\textasciicircum{}-14\textasciicircum{} \{\{ar\}\} \tn 
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\SetRowColor{LightBackground}
 & {[}H+{]}{[} 2×10\textasciicircum{}-3\textasciicircum{}{]} = 1 × 10\textasciicircum{}-14\textasciicircum{} \{\{ar\}\} \tn 
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\SetRowColor{white}
 & {[}H+{]} = 5.0 × 10\textasciicircum{}-12\textasciicircum{} M \{\{ar\}\} \tn 
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% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{2.{\bf{ hydroxide ion concentration in 0.002M HCl solution }}\{\{ac\}\}} \tn 
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\SetRowColor{white}
 & HCl (aq) → H+ (aq) + Cl- (aq) \{\{ar\}\} \tn 
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\SetRowColor{LightBackground}
{\emph{HCl is a strong acid, will dissociate 100\%}} & Kw = {[}H+{]}{[}OH-{]} = 1 × 10\textasciicircum{}-14\textasciicircum{} \{\{ar\}\} \tn 
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\SetRowColor{white}
 & {[}0.002M{]}{[}OH-{]} = 1 × 10\textasciicircum{}-14\textasciicircum{} \tn 
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 & {[}OH-{]} = 5.0 × 10\textasciicircum{}-12\textasciicircum{} \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
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\begin{tabularx}{8.4cm}{x{3.92 cm} x{4.08 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{8.4cm}}{\bf\textcolor{white}{CALCULATING pH OF STRONG ACID \& BASE}}  \tn
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\SetRowColor{LightBackground}
1. {\bf{pH of 0.10M Ba(OH)2 solution}} & Ba(OH)2  +   H2O -{}-{}-\textgreater{} Ba\textasciicircum{}2+\textasciicircum{}   +  2OH- \tn 
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I & 0.10M           \textbackslash{}\textbackslash{}\textbackslash{}              0                  0 \tn 
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C & -0.10            \textbackslash{}\textbackslash{}\textbackslash{}          +0.10        +2(0.10) \tn 
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E & 0M                 \textbackslash{}\textbackslash{}\textbackslash{}          0.10              0.20 \tn 
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pOH = -log{[}OH-{]}  = -log{[}0.20M{]} = 0.6990 & pH = 14.00-0.6990 = 13.30 \tn 
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\SetRowColor{LightBackground}
\mymulticolumn{2}{x{8.4cm}}{answer: {\emph{basic solution}}}  \tn 
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\end{tabularx}
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\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{ACID EQUILLIBRIUM CALCULATIONS (pH from Ka)}}  \tn
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\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{calculate pH of 1.0M Acetic acid soln, using approximations. Ka=1.8×10\textasciicircum{}-5\textasciicircum{} \{\{nl\}\} HC2H3O2 + H2O ⇄ H3O\textasciicircum{}+\textasciicircum{} + C2H3O2\textasciicircum{}-\textasciicircum{} \{\{ac\}\}} \tn 
\mymulticolumn{1}{x{8.4cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}{\bf{I}}  1.0 M \textbackslash{}\textbackslash{} 0M  0M \{\{ac\}\}\{\{nl\}\}{\bf{C}} -x  \textbackslash{}\textbackslash{}\textbackslash{}\textbackslash{} +x  +x\{\{nl\}\} {\bf{E}} 1-x \textbackslash{}\textbackslash{}\textbackslash{}\textbackslash{} x  x \{\{nl\}\} Ka = {[}H3O+{]}{[}C2H3O2-{]}/{[}HC2H3O2{]}\{\{nl\}\} Ka = x\textasciicircum{}2\textasciicircum{}/1.0 - x = 1.8 × 10\textasciicircum{}-5\textasciicircum{} \{\{nl\}\}√x\textasciicircum{}2\textasciicircum{} = √1.8 × 10\textasciicircum{}-5\textasciicircum{} \{\{nl\}\} {\bf{x = 4.24 × 10\textasciicircum{}-3\textasciicircum{}M}}\{\{nl\}\} \%ionization = {[}H3O+{]}eq/{[}HC2H3O2{]}org × 100\% \{\{nl\}\} =4.24×10\textasciicircum{}-3\textasciicircum{}M/1.0M ×100\% = {\bf{0.424\%}} {\emph{\textless{}5\% can neglect!}}\{\{nl\}\} {\bf{pH =}} -log{[}4.24×10\textasciicircum{}-3\textasciicircum{}{]} ={\bf{2.30}}} \tn 
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\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{} \tn 
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% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{calculate pH of 1.0×10\textasciicircum{}-5\textasciicircum{}M Acetic acid soln ({\emph{diluted concentration}})\{\{nl\}\}HC2H3O2 + H2O ⇄ H3O+ + C2H3O2- \{\{ac\}\}} \tn 
\mymulticolumn{1}{x{8.4cm}}{\hspace*{6 px}\rule{2px}{6px}\hspace*{6 px}{\bf{I}} 1.0×10\textasciicircum{}-5\textasciicircum{}M \textbackslash{}\textbackslash{}\textbackslash{}\textbackslash{} 0M  0M\{\{nl\}\}{\bf{C}} 1.0×10\textasciicircum{}-5\textasciicircum{}M-x \textbackslash{}\textbackslash{}\textbackslash{} +x +x\{\{nl\}\}{\bf{E}} 1.0×10\textasciicircum{}-5\textasciicircum{}M-x \textbackslash{}\textbackslash{}\textbackslash{}\textbackslash{} x   x\{\{nl\}\}Ka =x\textasciicircum{}2\textasciicircum{}/ 1.0×10\textasciicircum{}-5\textasciicircum{}M - x= 1.8×10\textasciicircum{}-5\textasciicircum{}M\{\{nl\}\}√x\textasciicircum{}2\textasciicircum{} = √1.8 × 10\textasciicircum{}-10\textasciicircum{}\{\{nl\}\} {\bf{x = 1.34× 10\textasciicircum{}-5\textasciicircum{}}}\{\{nl\}\} \%ionization = {[}H3O+{]}/{[}HC2H3O2{]} × 100\% \{\{nl\}\}={\bf{13\%}} {\emph{\textgreater{} 5\% can't neglect!}}\{\{nl\}\}x\textasciicircum{}2\textasciicircum{}/ 1.0×10-5M = 1.8×10\textasciicircum{}-5\textasciicircum{}M\{\{nl\}\}x\textasciicircum{}2\textasciicircum{} + 1.8×10\textasciicircum{}-5\textasciicircum{}x - 1.8×10\textasciicircum{}-10\textasciicircum{}=0\{\{nl\}\}1.8×10\textasciicircum{}-5\textasciicircum{}±√(1.8e\textasciicircum{}-5\textasciicircum{})\textasciicircum{}2\textasciicircum{} - 4(1)(-1.8×10\textasciicircum{}-10\textasciicircum{})/ 2(1)\{\{nl\}\} = 7.15×10\textasciicircum{}-6\textasciicircum{}M = {[}H3O+{]}\{\{nl\}\} {\bf{pH =}} -log{[}7.15×10\textasciicircum{}-6\textasciicircum{}M{]}= {\bf{5.14}}} \tn 
% Row Count 26 (+ 14)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
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% That's all folks
\end{multicols*}

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