Determine reaction order and reaction rate In a solution of $\ce{Fe^3+}$ the concentration was measured to be:
$$\begin{array}{c|c} 
t/\pu{min} &amp; \ce{[Fe^3+]}/\pu{\mu M} \\ \hline
10\ &amp; 238\\
20 &amp; 227\\ 
40\ &amp; 206\\ 
60\ &amp; 187 \\ 
80\ &amp; 169\\ 
100\ &amp; 154 \\
\hline
\end{array}$$
The decrease in $\ce{Fe^3+}$ concentration is due to the hydrolysis of the ion and precipitation of hydroxide. Calculate the reaction order and reaction rate.

I am having trouble understanding how to tackle this problem. I began by find the reaction formula:
$$ \ce{Fe^{3+} + H_2O &lt;=&gt; Fe(OH)^{2+} + H^+} $$
So this is a second order reaction. I then did:
$$\begin{array}{c|c|c} 
\text{time} &amp; \ce{[Fe^3+]} &amp; \text{H2O} &amp; \ce{[Fe(OH)^2+]} &amp; \ce{[H+]}\\ \hline
t=0 &amp; a\ &amp; a &amp; - &amp; -\\ 
t=t &amp; a-x &amp; b-x &amp; x &amp; x\\ 
t = \infty &amp; - &amp; b-a &amp; a &amp; a \\ 
\hline
\end{array}$$
But I am unsure if I should include water and the hydrogen ion, and also how I should continue if this is correct. Can someone please push me in the right direction here. Thank you!
EDIT
First order equation:
$$ \ln \frac{a}{a-x} = kt$$
Is a then equal to $238$ and $a-x$ is the initial value ($238$) minus the next value? e.g., $238-227 = 11$.
If I plot:
$$\begin{array}{c|c|c} 
t &amp; \ln \left (\frac{a}{a-x}\right) \\ \hline
10 &amp; 0\\ 
20 &amp; 3.07\\ 
40 &amp; 2.43\\ 
60 &amp; 2.53\\ 
80 &amp; 2.58 \\ 
100 &amp; 2.76\\ 
\hline
\end{array}$$
I get that the slope is $0.0168$ and that would be the rate of the reaction?
The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the  next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next,
$$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.
As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}]  $$
now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$
the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$
For more about pseudo-first order reactions you can check one of my previous answers.