Jump to content

stoichiometry problems....


rogue

Recommended Posts

I can't get the right answer for this question:

Calculate the concentration of anions in a solution prepared by dissolving 0.585 grams of nickel(II) bromide in sufficient water to produce 100.0 mL of solution.

 

any help/explaination would be great thanks

Link to comment
Share on other sites

Ignoring the autoionization of water, you're looking for [br-] in the equation:

NiBr2 <-> Ni++ + 2Br-

 

Since you know the amount of solution, you can calculate the molarity of NiBr2. This should be enough information to solve the equilibrium expression:

 

Keq = [Ni++][br-]2/[NiBr2]

Link to comment
Share on other sites

Nickel bromide dissolves completely in water, all bromides are very soluble in water with the exception of silver.

 

Thus NiBr2 ---> Ni2+ + 2Br -

 

Look at it this way. You are being asked only to calculate the concentration of anions, so it would be more useful to know the mass of bromide ions rather than the mass of nickel bromide. The ratio of bromide ions is the formula mass of bromine in NiBr2 over the whole formula mass of NiBr2.

 

<bromide ratio>

(79.9 * 2) / (58.7 + 79.9 + 79.9) = 159.8 / 218.5 = 0.731

 

0.731 * 0.585 = 0.427635 g

 

thus you have the mass of bromide dissolved, so to get the concentration you convert this mass into the number of moles then divide it by the volume of water. I get 0.0535 mol l-1 for bromide.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.