Asked by Mahtab

Ksp


The solutions at the two Pb electrodes of a concentration cell were prepared as follows:
Cell A: A mixture of 1.00 mL of 0.0500 M Pb(NO3)2 with 4.00 mL of 0.0500 M KX (the soluble potassium salt of an unspecified monovalent ion X-).
Some PbX2(s) precipitates.

Cell B: 5.00 mL of 0.0500 M Pb(NO3)2.
The cell potential was measured to be 0.05500 V at 25 °C.

By use of the Nernst equation, determine the concentration (M) of Pb2+ in the solution of Cell A.

Answers

Answered by Mahtab
Thanks I found the answer
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions