A55.0 ml aliquot of a 1.50 m solution is diluted to a total volume of 278 ml. a 139 ml portion of that solution is diluted by adding 155 ml of water. what is the final concentration? assume the volumes are additive. concentration:

0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

C2 = 0.30 M

The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

Then,

0.30x139 = C2x294

C2 = 0.14 M

The final concentration is 0.140 M

Explanation:

n₁=M₁.V₁ (First equation)

n₁=1.50 M

V₁=55 mL

M₂=n₂/V₂ (Second Equation)

V₂=278 mL

n₁=n₂

M₂=M₁.(V₁/V₂) (Third Equation)

n₃=M₃.V₃ (Forth Equation)

V₃=139 mL

M₃=M₂

n₃=M₁.(V₁/V₂).V₃ (Forth Equation)

M₄=n₄/V₄ (Fifth Equation)

n₄=n₃

M₄=M₁.(V₁/V₂).(V₃/V₄) (Sixth equation)

V₄=V₃+155 mL (Seventh Equation)

M₄=M₁.(V₁/V₂).(V₃/(V₃+155 mL))

M₄=1.50 M . (55mL / 278 mL) . ((139mL)/(139mL+155mL))

M₄=1.50 M . (55mL / 278 mL) . (139mL/294mL)

M₄=0.140 M

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