71.0 mL of a 1.30 M solution is diluted to a total volume of 238 mL. A 119 mL portion of that...

Question:

{eq}71.0\ mL {/eq} of a {eq}1.30\ M {/eq} solution is diluted to a total volume of {eq}238\ mL {/eq}. A {eq}119\ mL {/eq} portion of that solution is diluted by adding {eq}105\ mL {/eq} of water. What is the final concentration? Assume the volumes are additive.

Dilution

The process of adding water to a concentrated solution to make a dilute solution is known as dilution. This is governed by a simple equation 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.

The first cycle of dilution is given below to solve for the final concentration of the solution. That is,

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 1.30\ M {/eq} is the intial concentration,
• {eq}V_1 = 71.0\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 238\ 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{(1.30\ M)(71.0\ mL)}{238\ mL}\\ M_2 = 0.3878\ M {/eq}

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

Where,

• {eq}M_1 = 0.3878\ M {/eq} is the intial concentration,
• {eq}V_1 = 119\ mL {/eq} is the intial volume,
• {eq}M_2 {/eq} is the final concentration,
• {eq}V_2 = 119\ mL +105\ mL = 224\ 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{(0.3878\ M)(119\ mL)}{224\ mL}\\ \boxed{M_2 = 0.2060\ M} {/eq}