Question

A 55.0 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:

Answer 1

0.14 M

Step-by-step 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

Answer 2

The final concentration is 0.140 M

Step-by-step explanation:

• We have to calculate the moles of the first aliquot:

n₁=M₁.V₁ (First equation)

n₁=1.50 M

V₁=55 mL

• Now we have to calculate the concentration of the second solution knowing that the moles of the first aliquot (278 mL) and the moles of the second solution are the same:

M₂=n₂/V₂ (Second Equation)

V₂=278 mL

n₁=n₂

• If we substitute the first equation into the second one, we obtain the following:

M₂=M₁.(V₁/V₂) (Third Equation)

• The second aliquot (139 mL) has the same concentration as the second solution, so we need to calculate the moles:

n₃=M₃.V₃ (Forth Equation)

V₃=139 mL

M₃=M₂

• If we substitute the third equation into the forth one, we obtain:

n₃=M₁.(V₁/V₂).V₃ (Forth Equation)

• Now we have to calculate the concentration of the final solution, knowing that the moles of second aliquot are the same as the moles of the final solution:

M₄=n₄/V₄ (Fifth Equation)

n₄=n₃

• When we substitute the Forth Equation into the fifth one, we obtain:

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

• Now we have to remember that the volume of the final solution is:

V₄=V₃+155 mL (Seventh Equation)

• Now we substitute the seventh equation into the sixth one and we obtain:

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