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% Document Info
\author{fongrsy}
\pdfinfo{
  /Title (salt-preparation.pdf)
  /Creator (Cheatography)
  /Author (fongrsy)
  /Subject (Salt Preparation Cheat Sheet)
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    {\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{Salt Preparation Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{fongrsy} via \textcolor{DarkBackground}{\uline{cheatography.com/65383/cs/16915/}}}
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\noindent
\begin{multicols}{3}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}fongrsy \\
  \uline{cheatography.com/fongrsy} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 3rd September, 2018.\\
   Updated 3rd September, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\begin{document}
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% www.emerson.emory.edu/services/latex/latex_169.html
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Definition of Salt}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A salt is a compound formed when the hydrogen ion in an acid is replaced by a metallic ion or an ammonium ion. It is an ionic compound with high melting and boiling point. Soluble salts dissolve in water to form oppositely charged ions in aqueous solutions thus it can conduct electricity.% Row Count 6 (+ 6)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Solubility Table}}  \tn
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Hydrated Salts}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Water is present in the crystals of certain salts which gives the salt its crystalline properties. The water present is known as water of crystallization which is easily removed by heating. \newline % Row Count 4 (+ 4)
E.g: Copper(II) sulfate crystals (CuSO4.5H2O) \newline % Row Count 5 (+ 1)
Hydrated Salt (Upon Heating) \{\{fa-arrow-right\}\} Anhydrous Salt + Water% Row Count 7 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.28156 cm} x{2.2885 cm} x{1.00694 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Salt Preparation}}  \tn
% Row 0
\SetRowColor{LightBackground}
Method & Constituents & Salt Formed \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Reaction with Acid & Acid + Excess Insoluble Reactant & Soluble Salt \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Titration & Acid + Soluble Reactant & Soluble Salt \tn 
% Row Count 6 (+ 2)
% Row 3
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\seqsplit{Precipitation} & 2 Soluble Reactants & \seqsplit{Insoluble} Salt \tn 
% Row Count 8 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\SetRowColor{LightBackground}
\mymulticolumn{3}{x{5.377cm}}{Questions to consider: \newline 1. Is the salt soluble in water? \newline 2. Are the starting reactants soluble in water?}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}---}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Reaction of Excess Insoluble Reactant with Acid}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Method: \newline % Row Count 1 (+ 1)
Acid + Metal (Mg, Al, Zn, Fe) \newline % Row Count 2 (+ 1)
Acid + Insoluble Base (MgO, ZnO, CuO) \newline % Row Count 3 (+ 1)
Acid + Insoluble Carbonate (MgCO3, ZnCO3) \newline % Row Count 4 (+ 1)
The insoluble reactant used is in excess. This is to ensure that all the acid is used up. The excess reactant is filtered from the salt solution at the end of the reaction. The salt solution formed is then heated to evaporate some water away to obtain a saturated solution then cooled to allow crystallization to occur to produce the salt crystals. \newline % Row Count 11 (+ 7)
When is all the acid used up? \newline % Row Count 12 (+ 1)
1. When there is no more effervescence produced (Metal / Carbonate) \newline % Row Count 14 (+ 2)
2. When there is a residue of reactant left behind (Base) \newline % Row Count 16 (+ 2)
Only moderately reactive metals can be used. \newline % Row Count 17 (+ 1)
Steps: \newline % Row Count 18 (+ 1)
1. Add excess reactant into the acid and stir until the effervescence stops. \newline % Row Count 20 (+ 2)
2. Filter to remove excess reactants. Collect the filtrate (salt solution). \newline % Row Count 22 (+ 2)
3. Heat filtrate to obtain a concentrated salt solution. \newline % Row Count 24 (+ 2)
4. Test for saturation with a glass rod. \newline % Row Count 25 (+ 1)
5. Leave the solution to cool and crystalise. \newline % Row Count 26 (+ 1)
6. Filter to obtain crystals. \newline % Row Count 27 (+ 1)
7. Wash with cold distilled water and dry with clean filter paper.% Row Count 29 (+ 2)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Titration}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Method: \newline % Row Count 1 (+ 1)
Acid + Soluble Base / Alkali ( NaOH, KOH, NH3(aq) ) \newline % Row Count 3 (+ 2)
As both the reactants in this method are solutions, titration is used to add the exact volume of one reactant to the other. An indicator is used in the beginning to determine the volume of the reactants required. However, the experiment is repeated without the indicator so as to not contaminate the salt produced. \newline % Row Count 10 (+ 7)
1. Fill up a burette with dilute acid. Note the initial burette reading (V1 cm\textasciicircum{}3\textasciicircum{}). \newline % Row Count 12 (+ 2)
2. Pipette 25.0 cm\textasciicircum{}3\textasciicircum{} of aqueous alkali solution into a conical flask. \newline % Row Count 14 (+ 2)
3. Add 1−2 drops of methyl orange indicator to the solution. \newline % Row Count 16 (+ 2)
4. Add dilute acid slowly from the burette until the solution just turns orange. This is the end-point. \newline % Row Count 19 (+ 3)
5. Stop adding acid. Record the final burette reading (V2 cm\textasciicircum{}3\textasciicircum{}). \newline % Row Count 21 (+ 2)
Repeat the experiment: \newline % Row Count 22 (+ 1)
1. Pipette 25.0 cm\textasciicircum{}3\textasciicircum{} of alkali solution into a beaker. Do not add indicator. \newline % Row Count 24 (+ 2)
2. Add (V2 – V1) cm\textasciicircum{}3\textasciicircum{} of dilute acid from the burette. \newline % Row Count 26 (+ 2)
3. Heat to saturate the solution. \newline % Row Count 27 (+ 1)
4. Cool and crystallise to obtain the salt. \newline % Row Count 28 (+ 1)
5. Filter and dry.% Row Count 29 (+ 1)
} \tn 
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\begin{tabularx}{5.377cm}{X}
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\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Precipitation}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Method: \newline % Row Count 1 (+ 1)
Reaction with 2 aqueous solutions. \newline % Row Count 2 (+ 1)
2 Salt Solutions OR 1 Salt Solution \& 1 Dilute Acid \newline % Row Count 4 (+ 2)
In order to SEPARATE the insoluble salt from its reacting mixture, its starting reactants MUST BE soluble. This is to ensure that the insoluble salt produced can be filtered away from the reaction mixture. \newline % Row Count 9 (+ 5)
AB (aq) + CD (aq) → AD (s) + BC (aq) \newline % Row Count 10 (+ 1)
A\textasciicircum{}+\textasciicircum{} (aq) + D\textasciicircum{}-\textasciicircum{} (aq) → AD (s) \newline % Row Count 11 (+ 1)
Steps: \newline % Row Count 12 (+ 1)
1. Add solution AB to solution CD in a beaker. A white precipitate of AD forms. \newline % Row Count 14 (+ 2)
2. Filter to obtain the precipitate. \newline % Row Count 15 (+ 1)
3. Wash the precipitate with cold, distilled water to remove impurities. \newline % Row Count 17 (+ 2)
4. Leave the precipitate of AD to dry. \newline % Row Count 18 (+ 1)
E.g Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2NaNO3 (aq)% Row Count 20 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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% That's all folks
\end{multicols*}

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