# How do you balance Co(OH)_3 + HNO_3 -> Co(NO_3)_3 + H_2O?

Apr 5, 2017

$C o {\left(O H\right)}_{3}$ + $3 H N {O}_{3}$ $\rightarrow$ $C o {\left(N {O}_{3}\right)}_{3}$ + $3 {H}_{2} O$

#### Explanation:

You really have to know your complex ions for this one!

On the left side of the equation we see, initially, one nitrate group ($N {O}_{3}$) On the right, we have three. So that means on the left we need three nitrate groups which means three $H N {O}_{3}$ (aka nitric acids).

$C o {\left(O H\right)}_{3}$ + $3 H N {O}_{3}$ $\rightarrow$ $C o {\left(N {O}_{3}\right)}_{3}$ + ${H}_{2} O$ (not balanced)

Now, on the left side of the left side of the unbalanced equation we have six hydrogens, but on the right we only have two. We can fix that by putting a 3 coefficient in front of the water:

$C o {\left(O H\right)}_{3}$ + $3 H N {O}_{3}$ $\rightarrow$ $C o {\left(N {O}_{3}\right)}_{3}$ + $3 {H}_{2} O$

Now we see that all the hydrogens balance out on both sides (six and six), all the nitrate groups balance out (three and three), the cobalt is balanced (one and one), and the oxygen (not counting the oxygen we have with the nitrate groups we already counted!) is also balanced (three and three).

Notice what I did here: I treated the nitrate groups as distinct species and balanced them as a whole even though the nitrate group has both nitrogen and oxygen! If you think about it, you will probably realize that I could have done the same thing with the hydroxide groups--water is just a hydroxide group wiht an extra hydrogen.

Now try a few and see if treating the complex ions as individual species works for you.