# Concentrated HCI is 12 M. What is the molarity of an HCI solution prepared by dilluting 10 ml...

## Question:

Concentrated {eq}\rm HCI {/eq} is {eq}12\ M {/eq}. What is the molarity of an {eq}\rm HCI {/eq} solution prepared by dilluting {eq}10\ ml {/eq} concentrated {eq}\rm HCI {/eq} to a total volume of {eq}250\ ml {/eq}?

## Calculating New Molarity

The basic equation governing the dilution process is given as $$M_1V_1 = M_2V_2$$

where,

• {eq}M_1{/eq} is the intial concentration,
• {eq}V_1{/eq} is the intial volume,
• {eq}M_2{/eq} is the final concentration,
• {eq}V_2{/eq} is the final volume.

Note that the change in the molar concentration of the solution is proportional to the change in the volume of the solution. DIlution can be attained by adding water to a concentrated solution, thus making a diluted solution.

Use the equation below to solve for the final concentration of the solution. That is,

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 12\ M {/eq} is the intial concentration,
• {eq}V_1 = 10 \ mL {/eq} is the intial volume,
• {eq}M_2 = {/eq} is the final concentration,
• {eq}V_2 = 250\ mL {/eq} is the final volume.

Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,

{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(12\ M)(10\ mL)}{250\ mL}\\ \boxed{M_2 = 0.48\ M} {/eq}