Question 786889
As a chemist, I would advice you not to do that. The solution will get hot, may boil over violently, and you could get burnt. Luckily this is just a math problem, and science does not apply, so we can assume that it can be done and that volumes are additive, too.
 
{{{x}}}= amount of 55% muriatic acid added, in liters.
The amount of muriatic acid in that is {{{0.55x}}} (I won't even ask if it's in liters or kilograms).
The amount of muriatic acid in the 4 liters of 80% muriatic acid is
{{{0.80*4=3.20}}} (in the same L or kg units).
The total amount of muriatic acid in the final mixture will be
{{{0.55x+3.20}}}.
The final volume (in math class, if not in real life) will be
{{{4+x}}} liters.
Since those {{{4+x}}} liters of final solution will be 65% muriatic acid, we can also calculate the amount of muriatic acid in the final mixture as
{{{0.65(4+x)=2.6+0.65x}}}
So, {{{2.6+0.65x=0.55x+3.2}}}
 
Solving {{{2.6+0.65x=0.55x+3.2}}}:
{{{0.65x=0.55x+0.6}}}
{{{0.65x-0.55x=0.6}}}
{{{0.1x=0.6}}} and multiplying both sides by 10 (or dividing both by 0.1, same thing)
{{{highlight(x=6)}}}
So you are expected to add the 4L of 80% acid to {{{highlight(6L)}}} of 55% acid.