Question

A Ca ISE and a SCE were immersed in a standard containing 2.0x10" M Ca**. The cell potential was +0.2714 V. The two electrodes were then rinsed and transferred to a sample; the cell potential was +0.2837 V. Calculate the concentration of Caassuming that the Ca ISE has an ideal slope of 0.02958 V.

Answer 1

The potential of SCE electrode = 0.242 V

The potential of Ca ISE cell = 2.868 V - 0.02958 V * Log[Ca²⁺]

Now, E_cell = E_SCE - E_{Ca ISE}

i.e. 0.2837 = 0.242 - {2.868 V - 0.02958 V * Log[Ca²⁺]}

i.e. 0.02958 V * Log[Ca²⁺] = 2.9097

i.e. Log[Ca²⁺] = 0.0102

i.e. [Ca²⁺] = 1.0237 M