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Calculations involving solution preparation and Dilutions Calculations involving solution preparation and Dilutions

Calculations involving solution preparation and Dilutions - PowerPoint Presentation

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Uploaded On 2018-03-23

Calculations involving solution preparation and Dilutions - PPT Presentation

Solutions Expressed as percent The term percent is Greek for per 100 When concentration is expressed in terms of percent the numerator is the amount of solute and the denominator is 100 units of total solution ID: 661875

volume solution percent concentration solution volume concentration percent stock working naoh prepare 100 solute total solutions weight type units

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Slide1

Calculations involving solution preparation and DilutionsSlide2

Solutions Expressed as percent

The term

percent

is Greek for

per 100

. When concentration is expressed in terms of percent, the numerator is the amount of solute and the denominator is 100 units of total solution.

There are three types of percent expressions which vary in their units:

Type 1: Weight per volume (w/v) percent

. This percent is expressed as the weight of the solute (in grams) per 100ml of total volume.

Type 2: Volume percent (v/v).

In a percent by volume expression, both the amount of solute and the total volume are expressed in volume units (ml solute per 100 ml

total

volume)

Type 3: Weight percent (w/w)

is an expression of concentration in which the weight of solute is in the numerator and the weight of the total solution is in the denominator (grams of solute per 100 g of solution). This expression is rare.Slide3

Percent calculations

Type 1: (w/v)

How many grams of

NaCl

are required to prepare 250 ml of a 2.5% (w/v) solution of

NaCl

?

First, what is the concentration, and what does it mean?

__

2.5 g__

100 ml TV

Second, set up a proportion to calculate the amount of solute in the solution you want to prepare

=

___X___250 ml TV

Last, cross multiply to solve for X

(2.5 g)(250 ml) = (100 ml) (X)

Answer: X = 6.25 gSlide4

Percent calculations

Type 2: (v/v)

How would you prepare 300 ml of 70% isopropanol (v/v)?

First, what is the concentration, and what does it mean?

__

70 ml

iospropanol

__

100 ml TV

Second, set up a proportion to calculate the amount of isopropanol you need to prepare the solution with the requested total volume

=

____X___

300 ml TV

Cross multiply to solve for X

(70 ml)(300ml) = (X)(100ml)

X = 210 ml isopropanol

Don’t forget!

The question asks “how would you prepare” so you need to know how much water to add to the isopropanol to get 300ml to answer the question!

Add 210 ml isopropanol to 90 ml deionized waterSlide5

Preparing dilute solutions from concentrated solutions

The C

1

V

1

= C

2

V

2 EquationFrequently working solutions (to be use in experiments) are prepared from stock solutions (concentrated solutions)The

C1

V1

= C2

V2

Equation is used to determine the amount of stock solution needed to prepare a diluted working solution of a particular concentrationConcentrationstock X Volume

stock = Concentrationfinal X

VolumefinalA stock solution is sometimes written as an “

X” solution, where X mean “times”; how many times more concentrated the stock is than normal. The working concentration is always 1XSlide6

Using the C

1

V

1

=C

2V2 Equation

How much of a stock solution of NaOH with a concentration of 10N is required to prepare a 5N NaOH solution with a final volume of 50 ml?

First, find the following three values:

The stock concentration (C

1)The working concentration (C

2)The final (working) volume (V

2)

Second, plug in the values into the equation

(10N

NaOH

)(V

2

) = (5N)(50 ml)

Don’t get hung up on the units for concentration, they cancel out anyway

!

V

2

= 25 ml StockSlide7

A different way

How much of a stock solution of

NaOH

with a concentration of 10N is required to prepare a 5N

NaOH

solution with a final volume of 50 ml?

I don’t use the C1V1= C2V2 Equation. Instead I use this method……

First

, divide the stock concentration by the working concentration to get your dilution factor

Sound familiar?

10N NaOH 5N NaOH

=

2

Note once again that the units for concentration cancel out

Second

, divide the final volume of the working solution by the dilution factor to get the volume of stock (or the volume of 1 part)

50ml 2

=

25 ml

ANSWER

The reason I like this method is two fold. First, sometimes you are not given the working concentration, you are only given a dilution factor (remember the problems we did earlier in the semester) and you can’t use C

1

V

1

= C

2

V

2

. Second, I can do the above problem in my head, but I cannot use that equation without writing it down.