# Question #c8765

##### 1 Answer

The equilibrium concentration of **0.890 M**.

The first part of the problem was already answered here, check it out:

Now for the second part. This time, you start with **1.00 M** of **2.00 M** of **ICE table** to determine the equilibrium concentration of

**I**....1.00.............2.00...............0...................0

**C**....(-x)...............(-x)...............(+x)...............(+x)

**E**...1.00-x..........2.00-x...........x....................x

Your cuadratic equation will look like this

This equation will produce two solutions for **2.66** and **0.890**. The first solution will be eliminated, since it implies that the equilibrium concentrations of

This means that the equilibrium concentration of