75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that...

Question:

75.0 mL of a 1.20 M solution is diluted to a total volume of 248 mL. A 124-mL portion of that solution is diluted by adding 111 mL of water. What is the final concentration? Assume the volumes are additive.

Dilution

This process is commonly used by chemists to make a dilute solution. Also, this involved adding water to a concentrated solution to make a dilute solution. This is basically governed by $$M_1V_1 = M_2V_2$$

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.

The final concentration of the solution is determined using the equation below for the first cycle of dilution. That is,

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 1.2\ M {/eq} is the intial concentration,
• {eq}V_1 = 75.0\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 248\ 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{(1.20\ M)(75.0\ mL)}{248\ mL}\\ M_2 = 0.3629\ M {/eq}

Now consider the second cycle of dilution. Solving the final concentration using the same equation.

Where,

• {eq}M_1 = 0.3629\ M {/eq} is the intial concentration,
• {eq}V_1 = 124\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 124\ mL +111\ mL = 235\ 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{(0.3629\ M)(124\ mL)}{235\ mL}\\ \boxed{M_2 = 0.1915\ M} {/eq}