Question 774414
To do this properly, you really need to know the density or specific gravity of each silver iodiDe solution.  I assume you mean, "silver iodide", and not "silver iodine".  


If you want to ignore density, you could proceed like this:


Let v = volume of the 10% silver iodide solution to use.
{{{(8*4+v*10)/(8+v)=6}}}
{{{12+10v=6(v+8)}}}
{{{12+10v=6v+48}}}
{{{4v=36}}}
{{{v=9}}}
Nine liters of the 10% solution to add.