# Seeing solubility in various cases

[![enter image description here][1]][1] Question is in the image 'match' each option having multiple answers. I have invested a considerable time over this but couldn't solve it. Anyone explaining how to see the solubility in those cases will be appreciated. I am having confusion. Or you can give a hint to each given in column 2 or you can also suggest some specific reading. I tried this through reaction like agbr+h2o --->agoh+hbr now how do I know its solubility is greater or smaller than expectations!? Similar problems I am facing with other options. [1]: https://i.stack.imgur.com/0Yaot.jpg

• At the very least you may give only hints. – SM Sheikh Mar 31 at 11:12
• What things should I add to this to avoid it get 'closed'? – SM Sheikh Mar 31 at 11:25
• First, I suggest to improve the quality by getting rid of irrelevant stuff such as giant black margins and interface elements on the screenshot, e.g. crop and downscale the image. Or, better, use a MathJax table. If you don't know how to use MJ, let me know, I can do this one for you as an example. Second, feel free to add your attempt (a thought process), and what have you tried/read/analyzed to solve the problem. – andselisk Mar 31 at 11:29
• @andselisk i tried a different image but it shows 'an error occurred' – SM Sheikh Mar 31 at 11:39
• You are not using the mobile StackExchange app, are you? If you are, I strongly suggest to switch to the mobile web browser. The app hasn't been updated for years and has numerous issues at this point. My take on that — use a PC for content creation and a mobile phone for the notifications and content consumption. – andselisk Mar 31 at 11:41

A is wrong. $$\ce{AgBr}$$ is one of the least soluble compounds of $$\ce{Ag}$$. Why don't you have a look in the table of solubility products to be convinced? Further more, $$\ce{AgBr}$$ will never react with $$\ce{H_2O}$$ to produce a strong acid like $$\ce{HBr}$$ and a base like $$\ce{AgOH}$$. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water.
To solve D, you must know that $$\ce{Zn(OH)_2}$$ is soluble in an excess of $$\ce{OH^-}$$ ions