The solubility product, Ksp, of a sparingly soluble salt can be determined by measuring the cell potential of a cell known as a concentration cell - CIE - A-Level Chemistry - Question 2 - 2018 - Paper 1

One reason the voltmeter reading would change is that the concentration of Ag+(aq) can change if water evaporates from the cell solutions, thereby altering the internal concentration conditions and affecting the cell potential.

| 0.0010 | -2.00 | -0.097 | -1.56 | | 0.0050 | -1.30 | -0.140 | -1.74 | | 0.0100 | -1.00 | -0.159 | -1.79 | | 0.0500 | -0.60 | -0.174 | -1.10 | | 0.1000 | -0.20 | -0.202 | -0.96 | | 0.2000 | 0.00 | -0.221 | -0.76 | | 0.5000 | 0.30 | -0.239 | -0.56 | | 1.0000 | 0.70 | -0.264 | -0.33 | | 1.5000 | 1.00 | -0.294 | -0.12 |

To calculate the volume of Ag+(aq) needed, we can use the formula:

$C_1V_1 = C_2V_2$

Where:

• $C_1$ = 2.0 mol/dm³ (concentration to be prepared)
• $V_2$ = 250.0 cm³ (desired final volume)
• $C_2$ = desired concentration

Thus:

$2.0 \times V_1 = 0.250$ Solving for $V_1$, we get:

$V_1 = \frac{0.250}{2.0} = 0.125\text{ dm}^3 = 125\text{ cm}^3$

A suitable piece of apparatus to measure the calculated volume of Ag+(aq) would be a burette, as it allows for precise volume measurements.

Using the relationship $Ksp = Csat^2$ and assuming the concentration of saturated AgCl solution is equal to the calculated concentration, we can find Ksp:

Assuming Csat from part (c)(iii) gives us:

$Ksp = (12.5 \times 10^{-3})^2 = 1.56 \times 10^{-4}$

The graph is expected to curve upwards, starting off flat at low temperatures and gradually increasing as the temperature rises, indicating that Ksp increases with temperature.

Chloride ions can form a precipitate with silver ions (AgCl), disrupting the equilibrium and affecting the cell potential.