-3
$\begingroup$

enter image description here

why do we avoid the [AgCl] to calculate K?

$\endgroup$

closed as off-topic by Jan, NotEvans., Pritt Balagopal, DSVA, Mithoron Sep 9 '17 at 13:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

5
$\begingroup$

$\ce{AgCl}$ is a solid, and those are not included in the reaction quotient (or set to 1). It's handled e.g. here:

The activities of pure solids and liquids are equal to 1

Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1.

Because it's a fraction, you can leave out the 1.

$\endgroup$
-2
$\begingroup$

AgCl silver chloride is in solid state hence does not affect equilibrium.

$\endgroup$
  • 1
    $\begingroup$ Yeah, but it would help the OP (and future visitors) if you could briefly explain why it being a solid means it has negligible effects on equilibrium. $\endgroup$ – paracetamol Sep 9 '17 at 13:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.