I have one final question for this evening and this evening alone. I know it is Friday night so I don't expect a big return. Thanks to all who have helped me. This is my last chemisty problem of the quarter!
A solution is prepared by adding 0.075 mol of K3[Fe(CN)6] to 0.52 L of 2.0 M NaCN. Assuming no volume change, calculate the concentrations of Fe(CN)63- and Fe3+ in this solution. The K (overall) for the formation of Fe(CN)63- is 1 x 1042.
So, I know that because Kf is so large that essentially all of the Fe3+ exists within the solution as Fe(CN)63-. However, besides this information, I am a bit stumped. Thanks in advance to anyone who can *delete me*
EDIT: I figured it out!!