What volume of 4M HCl and 2M HCl should be mixed to get 500 mL of `2.5 M HCl` ?

Explanation

Let the volume of 4M HCl required to prepare 500 mL of `2.5 mL of 2.5M HCl =x ` mL
Therefore, the required volume of `2M HCl = (500-x)` mL
We know from the equation
`C _(1) V _(1) + C _(2) V_(2) = C_(3) V_(3)`
`(4x ) + 2 ( 500 -x) = 2.5 xx 500`
` 4x + 1000 - 2x = 1250`
`2x = 1250 - 1000`
`x = (250)/(2) = 125 mL`
Hence, volume of 4M HCl required = 125 mL
Volume of 2M HCl required `= (500 - 125) mL = 375 mL`