What is the colour of the resultant solution at the end point? Calculate the average volume of Q used.

The question refers to a titration where iron(II) ammonium sulphate is titrated with potassium dichromate(VI). Potassium dichromate(VI) is an oxidizing agent and is orange. Iron(II) ions are pale green. At the end point of this redox titration, the iron(II) ions have been completely oxidized to iron(III) ions (which are yellow/brown in solution). The first excess drop of dichromate will cause the solution to turn orange.

The resultant solution at the end point will be orange.

The volume of Q used is the difference between the final reading and the initial reading.

For the 1st titration:

$$\text{Volume of Q used} = 20.6\, \text{cm}^3 - 0.0\, \text{cm}^3 = 20.6\, \text{cm}^3$$

For the 2nd titration:

$$\text{Volume of Q used} = 20.0\, \text{cm}^3 - 0.0\, \text{cm}^3 = 20.0\, \text{cm}^3$$

For the 3rd titration:

$$\text{Volume of Q used} = 20.1\, \text{cm}^3 - 0.0\, \text{cm}^3 = 20.1\, \text{cm}^3$$

To calculate the average volume, we sum the concordant (close) readings and divide by the number of readings. In titration, concordant readings are usually within $\pm 0.1\, \text{cm}^3$ of each other. The readings are $20.6\, \text{cm}^3$, $20.0\, \text{cm}^3$, and $20.1\, \text{cm}^3$. The readings $20.0\, \text{cm}^3$ and $20.1\, \text{cm}^3$ are concordant. The reading $20.6\, \text{cm}^3$ is not concordant with the other two. Therefore, we use $20.0\, \text{cm}^3$ and $20.1\, \text{cm}^3$ for the average.

$$\text{Average volume of Q used} = \frac{20.0\, \text{cm}^3 + 20.1\, \text{cm}^3}{2}$$ $$\text{Average volume of Q used} = \frac{40.1\, \text{cm}^3}{2}$$ $$\text{Average volume of Q used} = 20.05\, \text{cm}^3$$

a) The colour of the resultant solution at the end point is orange.

b) The average volume of Q used is $\boxed{\text{20.05 cm}^3}$.