# Calculate the concentration of the reactant from the absorbance of the product without molar absorptivity

Given the reaction $$\ce{A->B}$$, where the absorbance of the product $$\ce{B}$$ has been measured (see the table below), and the initial concentration of $$\ce{A}$$ is $$\pu{10^{-4} M},$$ calculate the concentration of the reactant at the time $$t.$$

$$\begin{array}{rr} \hline t/\pu{min} & A \\ \hline 0 & 0.000 \\ 1 & 0.115 \\ 2 & 0.188 \\ 3 & 0.237 \\ 4 & 0.273 \\ 5 & 0.301 \\ 6 & 0.321 \\ 7 & 0.339 \\ 8 & 0.353 \\ 9 & 0.365 \\ 10 & 0.375 \\ \infty & 0.500 \\ \hline \end{array}$$

From the answer it is stated that one should use the following equation to calculate the concentration of $$\ce{A}$$:

$$[\ce{A}]_t = [\ce{A}]_0\left(1 – \frac{A_t}{A_\infty}\right)\label{eqn:1q}\tag{1}$$

I don't understand where this comes from since I have only ever converted absorbance to concentration using the equation:

$$A = \varepsilon lc\tag{2}$$

Could someone please explain the reasoning behind the equation \eqref{eqn:1q}?

• Stoichiometry dictates that the amount of A and B are related. You're expressing the concentration of A by taking the absorbance of B and relating that to the concentration of B.
– Zhe
May 23 at 13:00
• The equation can be written also as [A] = [A_0]-[B] = [A_0] - abs/abs_fin x [B_fin] May 23 at 13:18
• Because $A\to B$ final amount of B or $[B]_\infty=[A]_0=10^{-4}$, and as $Abs=\epsilon [C]l$ then $\epsilon l=0.5/10^{-4}$. At time $t,\; [A]_t=[A]_0-Abs_t/(\epsilon l)$. May 23 at 14:47
• I bet you must be aware of the general concept of proportionality ($y = a.x \implies y_2 = x_2 \cdot \frac{y_1}{x_1}$ without need to know $a$) and of constant sum of 2 variables ( $a = x + y \implies y = a - x$ ). Why have not you tried to apply it? May 24 at 7:01

Reactant $$\ce{A}$$ does not absorb the light, so the measured absorbance is only due to $$\ce{B}$$. Let $$[\ce{B}]_t$$ and $$A_t$$ be the measured values of $$[\ce{B}]$$ and $$A$$ at any time during the experiment:

$$A_t = \varepsilon_\ce{B} \cdot l \cdot [\ce{B}]_t \label{eqn:1a} \tag{1}$$

At the end of the experiment $$t = \infty$$, $$A_\infty = 0.500$$ and $$[\ce{B}]_\infty = [\ce{A}]_0 = \pu{E-4 M}$$. As a consequence

$$\varepsilon_\ce{B}\cdot l = \frac{A_\infty}{[\ce{B}]_\infty} = \frac{A_\infty}{[\ce{A}]_0} \label{eqn:2a}\tag{2}$$

Introducing \eqref{eqn:2a} in \eqref{eqn:1a} yields

$$[\ce{B}]_t = \frac{A_t}{\varepsilon_\ce{B}\cdot l} = [\ce{A}]_0 \frac{A_t}{A_\infty} \tag{3}$$

\begin{align} [\ce{A}]_t &= [\ce{A}]_0 - [\ce{B}]_t \\ &= [\ce{A}]_0 - [\ce{A}]_0 \frac{A_t}{A_\infty} \\ &= [\ce{A}]_0\left(1 – \frac{A_t}{A_\infty}\right) \tag{4} \end{align}

And this is the wanted formula.

• I apologize for the late reply, but how did you derive [A]$_t$ = [A]$_0$ - [B]$_t$ ? I don't see how you could get a subtraction between the two from eq. ($3$) May 27 at 15:17
• @Katara. Suppose [$\ce{A]_0] = 1.00} M$. If $30$% of this amount has been transformed into $\ce{B}$ after a time $t$, [$\ce{B]_t = 0.300} M$. Simultaneously [$\ce{A]_t = 1.00 - 0.30 = 0.70} M$ May 27 at 20:07
• Thank you very much! May 28 at 10:18