What volume of 9M HCl and 4 MHCl should be mixed to get 1L of 5M HCl solution?

Solution

Suppose volume of 9 MHCl required to prepare 1 L of 6M HCI = x litre

therefore

Volume of 4 M HCl required = (1 - x) litre
Applying molarity equation, we have,

stack straight M subscript 1 straight V subscript 1 with 9 space straight M space HCl below space plus stack space straight M subscript 2 straight V subscript 2 space with 4 space straight M space HCl below space equals space stack straight M subscript 3 straight V subscript 3 with 6 space straight M space HCl below

9 space cross times space straight x space plus 4 left parenthesis 1 minus straight x right parenthesis space equals space 6 cross times 1 or space space space space space 5 straight x space equals space 2 or space space space space space space straight x space equals space 2 over 5 space equals space 0.04 space litre therefore space space space Volume space of space 9 straight M space HCl space required space equals space 0.4 space litre space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 400 space mL therefore space space space space Volume space of space 4 straight M space HCl space required space equals space 1000 minus 400 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 600 space mL.

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