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Considering this as a mathematical problem rather than a chemical one, so not considering any on the contamination problems associated with diluting the acid, your solution should be made up 54% of your original acid and 46% water.
An easy way of working ot the percentage of the current solution that you need is to work in decimals and divide the required percentage of HCl by the current percentage of HCl. I have included the Mathematics of this below.
your current solution (call it solution A) is 37 parts hydrochloric acid and 63 parts water.
So the quantity of solution A can be described as
A = 0.37H + 0.63W
Your 20% solution (call it solution B) is 20 parts hydrochloric acid and 80 parts water.
So the quantity of B can be described as
B = 0.2H + 0.8W
However, B can also be descibed as an unknown percentage of A and the remaining percentage will be water
if the percentage of A written in decimal form is x
then the percentage of water is 1-x
so
B = Ax + (1-x)W
Equating the two equations for B gives:
0.2H + 0.8W = Ax + (1-x)W
Substituting in for A gives:
0.2H + 0.8W = x(0.37H + 0.63W) + (1-x)W
0.2H + 0.8W = 0.37xH + (0.63x +1 - x)W
Equating coefficents of H
0.2 = 0.37x
x = 0.2/0.37
x= 0.540540540 ...
x = 0.54 (2.s.f)
check this with equating W
0.8 = 0.63x + 1 - x
0.8 = 1 - x(1 - 0.63)
0.2 = x(1 - 0.63)
x = 0.2/(1 - 0.63)
x =0.540540540 ...
x = 0.54 (2.s.f)
so your solution should be made up 54% of your original acid and 46% water.
What else can I help you with?
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