Nickel(II) bromide plus potassium phosphate yields nickel (II) phosphate plus potassium br

Question

Write and balance the equation

Answer 1

The first step in all of these problems is to know the charges on the ions. You are given that the charge on nickel is +2 for both the nickel bromide reactant and the nickel phosphate product. You now need to know (or look up) the charges on the other ions. Bromide is Br (-1), potassium is K (+1) and phosphate is PO4 (-3). With that information, now you need to balance the charges on the individual components. For nickel bromide, in order to balance the +2 charge on the nickel, you would need two Br (-1) ions, so nickel(II) bromide is

NiBr2. Using the same approach, you can determine that potassium phosphate must be K3(PO4) and that potassium bromide must be KBr. In the case of nickel(II) phosphate, you have to note that the nickel has a +2 charge while the phosphate has a -3 charge. In order to balance the charge on this product, you will need three nickel ions for every two phosphate ions, for a formula of Ni3(PO4)2. Now that you know the makeup of all the reactants and products, you are ready to write a balanced chemical equation.

__. NiBr2 +. __ K3(PO4) ---->. __ Ni3(PO4)2 + __ KBr

Now you need to fill in the blanks for the coefficients. This can be done in a number of ways. I like to start with the material with the largest charge and go from there. The phosphate has the largest charge, -3. Note that, right now, there are two on the right but only one on the left. Put a provisional 2 in front of the potassium phosphate to get

__. NiBr2 + 2 K3(PO4) ---->. 1 Ni3(PO4)2 + __ KBr

Now, note that there are three nickels on the right but only one on the left, so put a provisional 3 in front of the nickel bromide to get

3 NiBr2 + 2 K3(PO4) ---->. 1 Ni3(PO4)2 + __ KBr

Now, see what you have left. The nickels are now balanced, and so are the phosphates. On the left hand side you have a total of six bromines and six potassiums, so as the last step you need to have six potassium bromides on the right. The final balanced equation is

3 NiBr2 + 2 K3(PO4) ----> 1 Ni3(PO4)2 + 6 KBr