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

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:

 

K_eq = [Ni⁺⁺][br^-]²/[NiBr₂]

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.

Thanks... that really helped me out