## Answers

The potential of SCE electrode = 0.242 V

The potential of Ca ISE cell = 2.868 V - 0.02958 V * Log[Ca^{2+}]

Now, E_{cell} = E_{SCE} - E_{Ca ISE}

i.e. 0.2837 = 0.242 - {2.868 V - 0.02958 V * Log[Ca^{2+}]}

i.e. 0.02958 V * Log[Ca^{2+}] = 2.9097

i.e. Log[Ca^{2+}] = 0.0102

i.e. **[Ca ^{2+}] = 1.0237 M**

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