这是我的代码。它显示维度过大错误。我花了很多时间试图解决这个问题，但没有成功。请帮帮我
\begin{tikzpicture}
\begin{axis}[
title={ln K$_{sp}$ vs 1/Temperature},
xlabel={1/Temperature (10$^{-3}$K)},
ylabel={ln K$_{sp}$ (10 $^{-5}$ mol/L)},
xmin=3, xmax=3.6,
ymin=-24, ymax=-19,
xtick={3.1,3.15,3.2,3.25,3.3,3.35,3.4,3.45,3.5},
ytick={-20,-20.5,-21,-21.5,-22,-22.5,-23},
legend pos=north east,
xmajorgrids=true,
ymajorgrids=true,
grid style=dashed,
]
\addplot [thick, domain=-23.5:-19.5,samples=100,]{-4.8878*x-5.4578};
\addplot[mark=*] coordinates {( 3.3557,-22.9305 )};
\addplot[mark=*] coordinates {( 3.3,-21.927 )};
\addplot[mark=*] coordinates {( 3.246,-21.234 )};
\addplot[mark=*] coordinates {(3.1949,-20.8286 )};
\addplot[mark=*] coordinates {( 3.1447,-20.7233 )};
\addplot[mark=*] coordinates {( 3.0959, -20.5410)};
\addplot[mark=*] coordinates {( 3.0487, -20.3178 )};
\end{axis}
\end{tikzpicture}
我甚至尝试将小数点后的数字减少到 2 位，但错误仍然存在
我的全部代码：
\documentclass[12pt]{article}
\usepackage{mhchem}
\usepackage{pgfplots}
\usepackage{amssymb}
\usepackage{gensymb}
\pgfplotsset{width=15cm,compat=1.9}
\title{{\huge \textbf{Experiment 3}} \\ DETERMINATION OF SOLUBILITY AND SOLUBILITY PRODUCT OF A SPARINGLY SOLUBLE SALT AT ROOM TEMPERATURE BY CONDUCTOMETRIC METHOD \\ {\Large Chemistry Lab CY19003} }
\author{Shaurya Goyal \\ 20NA10036 \\Section 2}
\date{\today}
\begin{document}
\maketitle
\newpage
\section{Introduction}
Solubility is the property of a solid, liquid or gaseous chemical substance called solute to dissolve in a solid, liquid or gaseous solvent. The solubility of a substance fundamentally depends on the physical and chemical properties of the solute and solvent as well as on temperature, pressure and presence of other chemicals (including changes to the $pH$) of the solution. The extent of the solubility of a substance in a specific solvent is measured as the saturation concentration, where adding more solute does not increase the concentration of the solution and begins to precipitate the excess amount of solute.\\
\vspace{0.2cm}
The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol $K_{sp}$.
\section{Formula}
\begin{enumerate}
\item The specific conductance ($\textbf{k}$) of the constituent ions of the sparingly soluble salt may be calculated as :
\begin{center}
$\textbf{k}_{\text{saturated solution}} - \textbf{k}_{\text{water}} = \textbf{k}_{\text{ion}}$\\
\vspace{0.2cm}
\end{center}
\item The molar conductance $\lambda$ is: \begin{center}
$\lambda = \frac{\textbf{k}_{\text{ion}}}{\text{c}}$ \\
\vspace{0.2cm}
where $c$ is the concentration of the electrolyte in $mol/L$ \\
\vspace{0.2cm}
\end{center}
\item For a saturated solution of a sparingly soluble salt, such as $BaSO_{4}$:
\begin{center}
$c = \left( \frac{1}{\lambda \degree _{\frac{1}{2} \text{Ba}^{2+}} \ + \lambda \degree _{\frac{1}{2} \text{SO} _{4}^{2-}}} \right) \times \textbf{k}_{\text{ion}} = \frac{\textbf{k}_{\text{ion}}}{\lambda \degree _{\text{BaSO}_{4}}} $\\
\vspace{0.2cm}
where $\textbf{k} _{\text{ion}}$ is the sum of the ionic conductances of all the ions in the test salt\\ and c is the concentration of the salt in mol/L
\end{center}
\newpage
\item For the sparingly soluble salt $A_{x}B_{y}$ in aqueous solution, the solubilization at fixed temperature and pressure can be written as an equilibrium reaction where equilibrium exists between the solid (undissolved salt) and the ions solution:
\begin{center}
\ce{A_{x}B_{y} (s) <=> xA^{y+}(aq) + yB^{x-}(aq)} \\
the equilibrium constant is given by $K_{\text{sp}} = [A^{y+}]^{x} [B^{x-}]^{y}$
\end{center}
\item The heat of solution is:
\begin{center}
$ln \ K_{s} = - \frac{\Delta H \degree _{sol}}{R} \left( \frac{1}{T} \right) + \frac{\Delta S \degree _{sol}}{R}$
\end{center}
\end{enumerate}
\section{Graph 1}
\begin{tikzpicture}
\begin{axis}[
title={Solubility of salt vs Temperature},
xlabel={Temperature \textcelsius},
ylabel={c (solubility) (10 $^{-5}$ mol/L)},
xmin=15, xmax=60,
ymin=0, ymax=5,
xtick={20,25,30,35,40,45,50,55,60},
ytick={1,1.5,2,2.5,3,3.5,4},
legend pos=north east,
xmajorgrids=true,
ymajorgrids=true,
grid style=dashed,
]
\addplot [thick,
domain=22:57,samples=100,]{0.1055*x-1.5875};
\addplot[mark=*] coordinates {( 25,1.05 )};
\addplot[mark=*] coordinates {( 30,1.73 )};
\addplot[mark=*] coordinates {( 35,2.45 )};
\addplot[mark=*] coordinates {( 40,3 )};
\addplot[mark=*] coordinates {( 45,3.16 )};
\addplot[mark=*] coordinates {( 50, 3.46)};
\addplot[mark=*] coordinates {( 55, 3.87 )};
\end{axis}
\end{tikzpicture}
\vspace{0.2cm}
\begin{center}
The room temperature is 25 \textcelsius.\\
\vspace{0.2cm}
The solubility at room temperature is $1.05 \times 10^{-5}$ mol/L \\ The $K_{sp}$ is $ 1.1 \times 10^{-10}$
\end{center}
\newpage
\section{Graph 2}
\begin{tikzpicture}
\begin{axis}[
title={ln K$_{sp}$ vs 1/Temperature},
xlabel={1/Temperature (10$^{-3}$K)},
ylabel={ln K$_{sp}$ (10 $^{-5}$ mol/L)},
xmin=3, xmax=3.5,
ymin=-24, ymax=-19,
xtick={3.1,3.15,3.2,3.25,3.3,3.35,3.4},
ytick={-23,-22.5,-22,-21.5,-21,-20.5,-20},
legend pos=north east,
xmajorgrids=true,
ymajorgrids=true,
grid style=dashed,
]
\addplot [thick, domain=-23.5:-19.5,samples=100,]{-4.8878*x-5.4578};
\addplot[mark=*] coordinates {( 3.35,-22.93 )};
\addplot[mark=*] coordinates {( 3.30,-21.92 )};
\addplot[mark=*] coordinates {( 3.24,-21.23 )};
\addplot[mark=*] coordinates {(3.19,-20.82 )};
\addplot[mark=*] coordinates {( 3.14,-20.72 )};
\addplot[mark=*] coordinates {( 3.09, -20.54)};
\addplot[mark=*] coordinates {( 3.04, -20.31 )};
\end{axis}
\end{tikzpicture}
\vspace{0.2cm}
\begin{center}
The slope is -4.8878 (mathematically).\\
\vspace{0.2cm}
$- \frac{\Delta H \degree}{R} = -4887.8$\\
$\therefore \Delta H \degree = 4887.8 \times 8.314$\\
$\therefore \Delta H \degree =40.64$ kJ/mol
\end{center}
\newpage