2BiOCl + NaOH = BiOH + NaCl. This is the reaction where I am trying to get Bismuth Hydroxide as one product and salt. My problem is what is the state of Bismuth Hydroxide? Whether it is BiOH or Bi (OH)2 or Bi(OH)3. How do I know which one I will get?


closed as off-topic by A.K., Jon Custer, Mithoron, Todd Minehardt, DrMoishe Pippik Aug 26 '18 at 22:40

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.

  • 2
    $\begingroup$ You know that from the valence of Bi, and that you deduce by looking at BiOCl. $\endgroup$ – Ivan Neretin Aug 24 '18 at 10:02

Well, the reaction actually produces $\ce{BiO.OH}$. According to this site:

It reacts with alkalis, especially when the solutions are concentrated. It is probable that with a dilute solution of potassium hydroxide a reversible reaction takes place according to the equation

$$\ce{BiOCl + KOH <=> BiO.OH + KCl}$$

With concentrated solutions the action is much more complex.

Some sources states that the reaction produces Bismuth(III) oxide. In conclusion, the reaction may produces different species depending on concentration of alkali, reaction conditions, amount of bismuth oxychloride etc.


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