# Acid - Base Equilibrium

Suppose we had a closed system proceeding according to,

$$\ce A +2\ce B \rightleftharpoons 2 \ce C$$

And that three experiments produced the following concentration data.

Experiment 1:

\begin{array}{|c|c|} \hline {\rm \small Species} & {\rm \small concentration} \\\hline \ce A & 0.500\rm~M \\ \ce B & 0.700\rm~M \\ \ce C & 0.900\rm~M \\\hline\end{array}

Experiment 2:

\begin{array}{|c|c|} \hline {\rm \small Species} & {\rm \small concentration} \\\hline \ce A & 0.300\rm~M \\ \ce B & 0.420\rm~M \\ \ce C & 0.424\rm~M \\\hline \end{array}

Experiment 3:

\begin{array}{|c|c|} \hline {\rm \small Species} & {\rm \small concentration} \\\hline \ce A & 1.40\rm~M \\ \ce B & {x}\rm~M \\ \ce C & 1.49\rm~M \\\hline \end{array}

How would you find the concentration of $\ce B$ from experiment 3 at equilibrium?

I assumed that each experiment was in equilibrium so that I could do the following,

$$K_c = \frac{\ce{[C]}^2}{[\ce{A}][\ce{B}]^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$

$$\therefore \frac{\ce{[C]}^2}{\ce{[A][B]}^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$

$$\therefore \ce{[B]} = \sqrt\frac{[1.490]^2[0.500][0.700]^2}{[0.900]^2[0.140]}$$

$$\therefore \ce{[B]} = 2.19\rm~M$$

However, $[\ce{B}] = 2.19\rm~M$ is not the answer. Could someone please provide some insight and direction into this question?

• Did you just ignore the experiment 2 altogether? – Ivan Neretin May 27 '16 at 10:31

Assuming you were supposed to interpolate $K_C$ from both experiment 1 and 2, you would do the following: Experiment1 yields $K_C=3.31$. Experiement two yields $K_C=3.40$. Interpolation yields $K_C=3.35$. This would give you $$B=\sqrt{\frac{C^2}{A*K_C}}=\sqrt{\frac{1.49^2}{1.4*3.35}}=0.69$$