Final answer:
To prepare 0.500 L of 0.250 M hydrochloric acid, approximately 0.071 L (or 71 mL) of 1.75 M hydrochloric acid must be diluted with water.
Step-by-step explanation:
To find the volume of 1.75 M hydrochloric acid that must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid, we can use the following formula:
M1V1 = M2V2
where M1 and V1 are the initial concentration and volume of the acid, and M2 and V2 are the final concentration and volume of the acid.
Plugging the known values into the equation, we get:
(1.75 M)(x) = (0.250 M)(0.500 L)
Solving for x, the volume of the 1.75 M hydrochloric acid that needs to be diluted, we find:
x = (0.250 M)(0.500 L) / (1.75 M)
x = 0.071 L
Therefore, approximately 0.071 L (or 71 mL) of 1.75 M hydrochloric acid must be diluted with water to prepare 0.500 L of 0.250 M hydrochloric acid.