MiKTeX and proTeXt ⇒ Error after MiKTeX Update

[localghost]

I ended up removing miketex from my computer and re-installing and then tried updating again and didn't get the error anymore. […]

The less time consuming method is to start the MiKTeX Package Manager (MPM), uninstall the l3kernel package and reinstall it again. This works for me and is done within a minute. The error disappeared.

[cbkschroeder]

Tafi,

Thanks for the step-by-step instructions; worked great!!

Christian

[localghost]

[…] Thanks for the step-by-step instructions; worked great! […]

The described process is not necessary.

[ghostanime2001]

There was another update to miktex yesterday and that update screwed up all the fonts when I ran XeLaTeX. All characters are wierd looking and some math symbols are even missing for example the multiplication symbol including the lines I drew with tikz. I have tried to to uninstall l3kernel, l3experimental, and even l3packages and reinstalled all 3. I have tried to re-install and still no help. This has happened to me yesterday as well and I re-installed yesterday, problem gone and late at night yesterday I tried to update hoping at the end of the day the author or whoever has fixed the problem by now and I updated. Checked this morning and it was all screwy. I don't get what's going on. Here is the full document which got screwed up. (no MWE's)

Code: Select all

\documentclass[fleqn]{article} 
\usepackage{amsfonts}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amstext}
\usepackage{amsthm}
\usepackage{array}
\usepackage{auto-pst-pdf}
\usepackage{booktabs}
\usepackage{bpchem}
\usepackage{calc}
\usepackage{cancel}
\usepackage{chemfig}
\usepackage{chemmacros}
\usepackage[shortlabels,inline]{enumitem}
\usepackage{fix-cm}
\usepackage[T1]{fontenc}
\usepackage{fouridx}
\usepackage{fullpage} 
\usepackage[margin=1in]{geometry} 
\usepackage{graphicx}
\usepackage[utf8]{inputenc}
\usepackage{lewis}
\usepackage{lmodern}
\usepackage{makebox}
\usepackage[version=3]{mhchem}
\usepackage{multienum}
\usepackage{multirow}
\usepackage{nccboxes}
\usepackage{pbox}
\usepackage{pgfkeys}
\usepackage{pifont}
\usepackage{pst-math}
\usepackage{pst-node}
\usepackage{pst-plot}
\usepackage{pstricks}
\usepackage{tensor}
\usepackage{tikz}
\usetikzlibrary{matrix,calc}
\usepackage{textcomp}
\usepackage{type1cm}
\usepackage{ulem}
\usepackage{wasysym}
\setlength{\parindent}{0pt}
\setlength{\parskip}{0pt}
\setlength{\mathindent}{0pt}
\newlist{longenum}{enumerate}{6}
\setlist[longenum,1]{label=\arabic*.}
\setlist[longenum,2]{label=\alph*)}
\setlist[longenum,3]{label=\alph*)}
\setlist[longenum,4]{label=\alph*)}
\setlist[longenum,5]{label=\alph*)}
\setlist[longenum,6]{label=\alph*)}
\newcommand{\since}{\raisebox{0.56pt}{\rotatebox[origin=c]{180}{$\Large\wasytherefore$}}}
\newcommand{\thus}{\Large{\wasytherefore}}
\newcommand{\s}{\par\medskip}
\newcommand{\ssa}{\\ \addlinespace[\smallskipamount]}
\newcommand{\msa}{\\ \addlinespace[\medskipamount]}
\newcommand{\bsa}{\\ \addlinespace[\bigskipamount]}
\newcommand{\fk}[2]{\mbox{\smash{#1}\vphantom{#2}}}
\allowdisplaybreaks
\pagestyle{empty}
\begin{document}
\hfill Calculations Based on Bronsted-Lowry Theory\hfill\llap{} \\
\begin{longenum}[{A}-1.,nosep,leftmargin=0in]
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item \ce{HNO2}
\item \ce{H2CO3}
\item \ce{H2PO4-}
\item \ce{H2O}
\item \ce{H3SO4+}
\item \ce{CH3NH3+}
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item \ce{F-}
\item \ce{CO3^{2-}}
\item \ce{NH2-}
\item \ce{N2H4}
\item \ce{PO4^{3-}}
\item \ce{(CH3)2NH}
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{CO3^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{HF}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{base}\end{tabular}}}{\ce{F^{-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{NH4+}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{B}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{C}{D} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{CO3^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{HF}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{base}\end{tabular}}}{\ce{F^{-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{HCO3-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{D}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{C} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{HTe-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{acid}\end{tabular}}}{\ce{H3PO4}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{H2Te}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{base}\end{tabular}}}{\ce{H2PO4-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{acid}\end{tabular}}}{\ce{HCO3-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{base}\end{tabular}}}{\ce{S^{2-}}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{base}\end{tabular}}}{\ce{CO3^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{HS-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{H-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{acid}\end{tabular}}}{\ce{H2O}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{H2}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{base}\end{tabular}}}{\ce{OH-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{acid}\end{tabular}}}{\ce{H2Se}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{base}\end{tabular}}}{\ce{HO2-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{base}\end{tabular}}}{\ce{HSe-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{H2O2}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{O^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{acid}\end{tabular}}}{\ce{H2O}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{base}\end{tabular}}}{\ce{OH-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{OH-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{D}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{C} \s\s\s
\item
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{H2O}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{acid}\end{tabular}}}{\ce{H2SO3}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{H3O+}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{base}\end{tabular}}}{\ce{HSO3-}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D}
\end{longenum} \s\s\s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}ll@{}}\ce{Te^{2-} + H2O <=> HTe- + OH-} &\quad \begin{aligned}[t]\ce{K_{A}}&=\num{1.0e-11} \\ \ce{K_{B}}&=0.001\end{aligned}\end{array}$ \s
\item 
$\begin{array}[t]{@{}ll@{}}\ce{N2H5+ + H2O <=> N2H4 + H3O+} &\quad \ce{K_{A}}=\num{5.88e-9}\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}\ce{HS- + H2O <=> H2S + OH-} &\quad \ce{K_{B}}=\num{1e-7}\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}\ce{SO4^{2-} + H2O <=> HSO4- + OH-} &\quad \ce{K_{B}}=\num{7.69e-13}\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}\ce{Be^{2+} + 2H2O <=> BeOH+ + H3O+} &\quad \ce{K_{A}}=\num{2e-4}\end{array}$
\end{longenum} 
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item 
$\begin{array}[t]{@{}ll@{}}
\ce{H2Te <=> HTe- + H+} &\quad \ce{K_{A}}=\num{2.3e-3} \ssa
\ce{CH3COOH <=> CH3COO- + H+} &\quad \ce{K_{A}}=\num{1.76e-5} \msa
\ce{K_{A}$\left(\ce{H2Te}\right)$}>\ce{K_{A}$\left(\ce{CH3COOH}\right)$} \ssa
\text{$\thus$ \ce{H2Te} is the stronger acid}
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{H2O2 <=> HO2- + H+} &\quad \ce{K_{A}}=\num{2.4e-12} \ssa
\ce{HSO3- <=> SO3^{2-} + H+} &\quad \ce{K_{A}}=
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{H2PO4- <=> HPO4^{2-} + H+} &\quad \ce{K_{A}}=\num{6.3e-8} \ssa
\ce{HS- <=> S^{2-} + H+} &\quad \ce{K_{A}}=\num{1.3e-13} \msa
\ce{K_{A}$\left(\ce{H2PO4-}\right)$}>\ce{K_{A}$\left(\ce{HS-}\right)$} \ssa
\text{$\thus$ \ce{H2PO4-} is the stronger acid}
\end{array}$
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}ll@{}}
\ce{HCO3- <=> H2CO3 + OH-} &\quad \ce{K_{B}}=\num{2.1e-4} \ssa
\ce{Te^{2-} <=> HTe- + OH-} &\quad \ce{K_{B}}=0.001 \msa
\ce{K_{B}$\left(\ce{Te^{2-}}\right)$}>\ce{K_{B}$\left(\ce{HCO3-}\right)$} \ssa
\text{$\thus$ \ce{Te^{2-}} is the stronger base}
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{HPO4^{2-} <=> H2PO4- + OH-} &\quad \ce{K_{B}}=\num{2.3e-2} \ssa
\ce{HS- <=> H2S + OH-} &\quad \ce{K_{B}}=\num{7.7e-2} \msa
\ce{K_{B}$\left(\ce{HPO4^{2-}}\right)$}>\ce{K_{B}$\left(\ce{HS-}\right)$} \ssa
\text{$\thus$ \ce{HPO4^{2-}} is the stronger base}
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{OH- <=> H2O + OH-} &\quad \ce{K_{B}}= \ssa
\ce{NH3 <=> NH4+ + OH-} &\quad \ce{K_{B}}=\text{very large} \msa
\text{\ce{NH3} is the stronger base}
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{HSe- <=> H2Se + OH-} &\quad \ce{K_{B}}= \ssa
\ce{HSO3- <=> H2SO3 + OH-} &\quad \ce{K_{B}}=
\end{array}$
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}l@{}}
\ce{Na2SO3 + H2O <=> 2Na+ + OH-} \ssa
\ce{SO3^{2-} + H2O <=> HSO3- + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{K2Te + H2O <=> 2K+ + Te^{2-}} \ssa
\ce{Te^{2-} + H2O <=> HTe- + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{Al(H2O)5(OH)Cl2 + H2O <=> Al(H2O)5(OH)^{2+} + Cl^{2-}} \ssa
\ce{Al(H2O)5(OH)^{2+} + H2O <=> Al(H2O)_{6}^{3+} + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{Na2HPO4 + H2O <=> 2Na+ + HPO4^{2-}} \ssa
\ce{HPO4^{2-} + H2O <=> H2PO4- + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{LiHCO3 + H2O <=> Li+ + HCO3-} \ssa
\ce{HCO3- + H2O <=> H2CO3 + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{K2O + H2O <=> 2K+ + O^{2-}} \ssa
\ce{O^{2-} + H2O <=> OH- + OH-}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{LiBr + H2O <=> Li+ + Br-} \\
\text{No hydrolysis reaction}
\end{array}$ \\
\item
$\begin{array}[t]{@{}l@{}}
\ce{K2C2O4 + H2O <=> 2K+ + C2O4^{2-}} \ssa
\ce{C2O4^{2-} + H2O <=> HC2O4- + OH-}
\end{array}$
\end{longenum} \par\smallskip
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}l@{}}
\ce{NaNO2 + H2O <=> Na+ + NO2-} \ssa
\ce{NO2- + H2O <=> HNO2 + OH-} \ssa
\text{basic}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{NH4Br + H2O <=> NH4+ + Br-} \ssa
\ce{NH4+ + H2O <=> NH3 + H+} \ssa
\text{acidic}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{Cr(H2O)6Cl3 + H2O -> Cr(H2O)6^{3+} + 3Cl-} \ssa
\ce{Cr(H2O)6^{3+} + H2O <=> Cr(H2O)6(OH)^{2+} + H+} \ssa
\text{acidic}
\end{array}$
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}l@{}}
\begin{aligned}[t]\ce{K_{A}$\left(1\right)$}\times\ce{K_{A}$\left(2\right)$}&=\ce{K_{total}} \\ \left(\num{1e-7}\right)\left(\num{1.3e-13}\right)&=\num{1.3e-20}\end{aligned}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\begin{aligned}[t]\ce{K_{A}$\left(1\right)$}\times\ce{K_{A}$\left(2\right)$}\times\ce{K_{A}$\left(3\right)$}&=\ce{K_{total}} \\ \left(\num{7.3e-10}\right)\left(\num{1.8e-13}\right)\left(\num{1.6e-14}\right)&=\num{2.1e-36}\end{aligned}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\begin{aligned}[t]\ce{K_{A}$\left(1\right)$}\times\ce{K_{A}$\left(2\right)$}&=\ce{K_{total}} \\ \left(\num{3.5e-3}\right)\left(\num{5e-8}\right)&=\num{1.8e-10}\end{aligned}
\end{array}$
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2S}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{NH3}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HS-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{NH4+}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{H2S}\right)$}}{\ce{K_{A}$\left(\ce{NH4+}\right)$}}=\dfrac{\num{1e-7}}{\num{5.59e-10}}=180
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2PO4-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HS-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HPO4^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2S}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{H2PO4-}\right)$}}{\ce{K_{A}$\left(\ce{H2S}\right)$}}=\dfrac{\num{6.3e-8}}{\num{1e-7}}=0.63
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{NH4+}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{OH-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{NH3}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2O}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{NH4+}\right)$}}{\ce{K_{A}$\left(\ce{H2O}\right)$}}=\dfrac{\num{5.59e-10}}{\num{1e-14}}=56000
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2O2}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{SO3^{2-}}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HO2-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HSO3-}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{H2O2}\right)$}}{\ce{K_{A}$\left(\ce{HSO3-}\right)$}}=\dfrac{\num{2.4e-12}}{\num{6.3e-8}}=\num{3.8e-5}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{CH3COOH}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{PO4^{3-}}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{CH3COO-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HPO4^{2-}}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{CH3COOH}\right)$}}{\ce{K_{A}$\left(\ce{HPO4^{2-}}\right)$}}=\dfrac{\num{1.76e-5}}{\num{4.4e-13}}=40,000,000
\end{array}$ 
\item
$\begin{array}[t]{@{}l@{}}
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{CO3^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{2H2O}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2CO3}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{2OH-}}$} \ssa
\dfrac{\ce{K_{A}$\left(\ce{H2O}\right)$}}{\ce{K_{A}$\left(\ce{H2CO3}\right)$}}=\dfrac{\num{1e-28}}{\num{2.068e-17}}=\num{4.8e-12} \bsa
\begin{array}[t]{@{}l@{}ll@{}}
&\ce{H2O <=> H+ + OH-} &\quad \begin{aligned}[t]\ce{K_{A}$\left(1\right)$}=\num{1e-14}\end{aligned} \ssa
&\ce{H2O <=> H+ + OH-} &\quad \begin{aligned}[t]\ce{K_{A}$\left(2\right)$}=\num{1e-14}\end{aligned} \\ \cmidrule{1-2}
&\ce{2H2O <=> 2H+ + 2OH-} &\quad \begin{aligned}[t]\ce{K_{total}}&=\ce{K_{A}$\left(1\right)$}\times\ce{K_{A}$\left(2\right)$} \\ &=\left(\num{1e-14}\right)\left(\num{1e-14}\right) \\ &=\num{1e-28}\end{aligned}
\end{array}
\end{array}$
\end{longenum} \s
\item
\begin{longenum}[nosep,widest=,align=left,leftmargin=*]
\item
$\begin{array}[t]{@{}l@{}}
\ce{NaHSO4_{(aq)} -> Na+ + HSO4-} \\
\ce{NaNO2_{(aq)} -> Na+ + NO2-} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HSO4-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{NO2-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{SO4^{2-}}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HNO2}}$} \ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{HSO4-}\right)$}}{\ce{K_{A}$\left(\ce{HNO2}\right)$}}=\dfrac{\num{1.25e-2}}{\num{5.1e-4}}=25 \ssa
\ce{K_{eq}}>1 \ssa
\text{$\thus$ products favoured}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{NH4F -> NH4+ + F-} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{NH4+}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{F-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{NH3}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HF$\vphantom{\ce{H2}}$}}$} \ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{NH4+}\right)$}}{\ce{K_{A}$\left(\ce{HF}\right)$}}=\dfrac{\num{5.8e-10}}{\num{6.7e-4}}=\num{8.7e-7}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{H3PO4 -> H2PO4- + H+} \ssa
\ce{Na2HPO4 <=> 2Na+ + HPO4^{2-}} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H3PO4}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HPO4^{2-}}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{H2PO4-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2PO4-}}$} \ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{H3PO4}\right)$}}{\ce{K_{A}$\left(\ce{H2PO4-}\right)$}}=\dfrac{\num{7.1e-3}}{\num{6.3e-8}}=\num{1.1e5}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{H2O2 -> HO2- + H+} \ssa
\ce{KHS <=> K+ + HS-} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2O2}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HS-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{HO2-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{H2S}}$} \ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{H2O2}\right)$}}{\ce{K_{A}$\left(\ce{H2S}\right)$}}=\dfrac{\num{2.4e-12}}{\num{1e-7}}=\num{2.4e-5}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{(NH4)2S -> 2NH4+ + S^{2-}} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{NH4+}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{S^{2-}}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}base\end{tabular}}}{\ce{NH3}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}acid\end{tabular}}}{\ce{HS-}}$} \ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{NH4+}\right)$}}{\ce{K_{A}$\left(\ce{HS-}\right)$}}=\dfrac{\num{5.8e-10}}{\num{1.3e-13}}=\num{4.46e3}
\end{array}$ \s
\item
$\begin{array}[t]{@{}ll@{}}
\ce{MgO -> Mg^{2+} + O^{2-}} &\quad \ce{K_{sp}}= \ssa
\mathrlap{\ce{O^{2-} + H2O <=> OH- + OH-}}
\end{array}$ 
\item
$\begin{array}[t]{@{}l@{}}
\ce{H2S <=> HS- + H+} \ssa
\ce{LiNO2 -> Li+ + NO2-} \ssa
\ce{NO2- + H2O <=> HNO2 + OH-} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{acid}\end{tabular}}}{\ce{H2S}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{base}\end{tabular}}}{\ce{NO2-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{base}\end{tabular}}}{\ce{HS-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{acid}\end{tabular}}}{\ce{HNO2}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \ssa\ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{H2S}\right)$}}{\ce{K_{A}$\left(\ce{HNO2}\right)$}}=\dfrac{\num{1e-7}}{\num{5.1e-4}}=\num{1.96e-4}\approx\num{2e-4}
\end{array}$ \s
\item
$\begin{array}[t]{@{}l@{}}
\ce{NaHCO3 -> Na+ + HCO3-} \ssa
\ce{NaHSO3 -> Na+ + HSO3-} \msa
\ce{$\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{A}{base}\end{tabular}}}{\ce{HCO3-}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{B}{acid}\end{tabular}}}{\ce{HSO3-}}$ <=> $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{C}{acid}\end{tabular}}}{\ce{H2CO3}}$ + $\underset{\clap{\begin{tabular}[t]{@{}c@{}}\Rnode{D}{base}\end{tabular}}}{\ce{SO3^{2-}}}$}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=0.5em]{A}{C}
\ncbar[linewidth=0.1pt,angle=-90,nodesep=3pt,arm=1.1em]{B}{D} \ssa\ssa
\ce{K_{eq}}=\dfrac{\ce{K_{A}$\left(\ce{HSO3-}\right)$}}{\ce{K_{A}$\left(\ce{H2CO3}\right)$}}=\dfrac{\num{6.2e-8}}{\num{4.4e-7}}=0.14
\end{array}$
\end{longenum} \s
\item
$\begin{array}[t]{@{}l@{}}
\begin{array}[t]{@{}ll@{}}
\begin{array}[t]{@{}c|c|c|c@{}}
& \ce{H2S <=>} & \ce{H+ +} & \ce{HS-} \\[1.5pt] \hline
\ce{S} & 0.05 & 0 & 0 \\
\ce{\Delta} & -x & +x & +x \\
\ce{E} & 0.05-x & x & x 
\end{array} &\quad \ce{K_{A}}=\num{1e-7}
\end{array} \msa
\begin{aligned}[t]&\ce{K_{A}}=\dfrac{\bigl[\ce{H+}\bigr]\bigl[\ce{HS-}\bigr]}{\bigl[\ce{H2S}\bigr]} \\ &\num{1e-7}=\dfrac{x^2}{0.05-x} \\ &\num{1e-7}=\dfrac{x^2}{0.05} \\ &0.05\left(\num{1e-7}\right)=x^2 \\ &\sqrt{0.05\left(\num{1e-7}\right)}=x \\ &\num{7.1e-5}=x\end{aligned}
\end{array}$ \s
\item
\noindent $\begin{array}[t]{@{}ll@{}}
\begin{tikzpicture}[baseline=(m-1-1.base)]
\matrix (m) [matrix of math nodes,ampersand replacement=\&]
{
\vphantom{\ce{H2S <=>}} \& \node (a) {\jvbox{\ce{H+ +}}[b]{\ce{H2S <=>}}}; \& \node (b) {\jvbox{\ce{H+ +}}[b]{\ce{H+ +}}}; \& \node (c) {\jvbox{\ce{H+ +}}[b]{\ce{HS-}}}; \\ \hline
\ce{S} \& \node {0.05}; \& \node {0}; \& \node {0}; \\
\Delta \& \node {-x}; \& \node {+x}; \& \node {+x}; \\
\ce{E} \& \node (d) {\smash[b]{\makebox*{\ce{H2S <=>}}{$0.05-x$}}}; \& \node (e) {\smash[b]{\makebox*{\ce{H+ +}}{$x$}}}; \& \node (f) {\smash[b]{x}}; \\
};
\draw (a.north west) -- (d.south west);
\draw (a.north east) -- (d.south east);
\draw (b.north east) -- (e.south east);
\end{tikzpicture}
\end{array}$
\end{longenum} 
\end{document}