Question 778238
Give a variable to how much pure acid.
Let y = volume in mL of pure acid to add.


Note that amount of acid in mixture or resulting solution will be {{{600*0.5+y*1}}} and volume of resulting solution will be {{{y+600}}} mL.  We must assume you mean your concentration as volume/volume; otherwise, you will need the densities of your pure acid and of the 50% solution.  


In your simplified description, your equation is {{{highlight((600*0.5+y)/(y+600)=0.75)}}};
{{{(300+y)=0.75(y+600)}}}
{{{y+300=0.75y+(3/4)600}}}
{{{y+300=0.75y+450}}}
{{{(1/4)y=150}}}
{{{y=4*150}}}
{{{highlight(y=600)}}} mL of the pure acid.