Re: What is your approach?

[Guest guest]

Dear Janet and all,

Well to begin with let us agree that there are many ways to skin a

cat (even if I love cats! and wouldn't if you paid me!) So what

ever method I use here on the site is not necessarily the method

that someone else would use and I am sure we will see other's

methods. Also I may even use a different metod ON the JOB because I

have a better understanding of math or shall we say 'mastered' the

math and therefore I may use a short cut. However my job on this

site is to explain and teach not just to give answers. So I am using

here that which I would use in my classroom:

1. A drug label states the addition of 8 ml of saline will result

in a drug concentration of 250mg/ ml. You realized that you added

13ml to the vial. What is the final concentration in the vial?

This problem must be worked backwards.

The following equation is used:

Strength

Concentration = ---------------

Total volume

Strength

Final Concentration = ---------------

1 ml volume

However in order to do this you would have to know the original

strength of the drug in the vial, such as 1 gram or 2 gram, Unless

you assume that there is NO powder volume.

Assuming that there is no displacement of the drug, known as powder

volume then the rest is easy:

If adding 8ml yields 250mg/1ml then there must have been 2 grams in

the vial to begin with. Proof:

250mg……………..X mg

------ = -------

1 ml………………….8ml

Cross multiply = 8 x 250mg/1 = 2000mg or 2 gram

Therefore as stated above we needed to know the original strength in

the vial. We NOW know it. It is 2 grams. Remember this is ONLY true

if the powder volume is ZERO (0).

Using that strength of 2 grams we can plug in the rest. If you

added 13 ml to th e 2grams with NO powder volume:

Strength

Concentration = ---------------

1 ml volume

2 grams

C = ----------

13 ml

The final volume is over 1ml.

Strength

Final Concentration = ---------------

1 ml volume

So we must set the two ratios equal to each other OR divide the

numerator (top) by the denominator (bottom).

2 grams………….X grams

------- = ---------

13 ml…………………1ml

cross multiply : 2gm x 1 /13 = 2 gram/13ml

Convert the 2 grams to 2000mg it now reads 2000mg/13ml

Division yields: 153.85 mg/ml

The new Final Concentration is: 153.85 mg/ml

Time it took to do this problem on the computer: 20 minutes

On the job it would take me about 2 -5 minutes.

2. Pharmacy has 300 ml of a 50% sol. 200 ml is added to this sol.

to decrease the concentration. How many grams of active ingredients

would there be in 4 oz of this diluted sol?

A. Recall from my tutorials that the word `of' between two numbers =

multiplication or " x " .

B. Well to begin the new volume of the diluted solution is 300ml +

200ml = 500 ml.

C. Next we only want to know about the number of grams in 4 ounces

of the 500 ml of diluted solution. Let's convert the 4 ounces to

ml: 4oz x 30ml/1oz = 120 ml.

D. Using the following formula: V1 x C1 = V2 x C2, where V – Volume

and C = concentration, we can plug in the numbers:

V1 x C1 = V2 x C2

300 ml x 50% = 500 ml x C2

300 ml x 0.5 = 500 ml x C2

300 ml x 0.5

------------- = C2

500 ml

0.3 = C2

Convert to percent multiply by 100 0.3 x 100 = 30% .

This means the new solution is a 30% solution.

Now what about the number of grams in 4 ounces (120 ml) of the new

solution?

Recall from my tutorials that X% = X grams/100 ml

Therefore 30% = 30 grams/100ml

Using ratio proportion:

30 grams……………X grams

--------- = ---------

100 ml……………………120 ml

Cross multiply:

30 grams x 120 ml / 100 ml = 36 grams

Therefore the number of grams in 4 ounces (120ml) of the new diluted

solution.

Above took 12 minutes to place on the post/computer .

Probably take me 2 -4 minutes.

3. The tech must weigh out 20 grams of menthol and is told to

dissolve it in sterile water to make a 5% solution. The RPH then

receives an order to change the concentration to a 1% sol. What

will be the final volume of the 1% solution.

Lets begin with knowing more about the 5% solution:

5% = 5 grams/100ml

But we used 20 grams! So set up the two ratios and multiply to solve

for the amount of volume using 20 grams:

5 grams…………..20grams

-------- = ---------

100 ml………………….Xml

Cross multiply:

100 x 20 / 5 = 400ml

Total volume is 400 ml with 20 grams of solution.

Now to make a 1% from the 5% let's use the formula from the previous

math problem:

V1 x C1 = V2 x C2

400 ml x 5% = 1% x C2

400ml x 0.05 = 0.01% x C2

400ml x 0.05

--------------- = C2

0.01

C2 = 2000 ml or C2 = 2 Liters

Therefore the final volume is 2 Liters

Above took 8 minutes

OJT about 2 - 4 minutes

4. You have D5W, D10W, and D20W in stock. What quantity of any two

solutions should be used to make 750 ml of a D12W solution?

First of all you can not get blood from a turnip! Or you can't make

a silk purse out of a sow's ear!

So you can not make a weak solution stronger by adding a weaker

solution!!

What does that mean?

It means you will never get a D12W or 12% Dextrose solution from

diluting a D10W with a D5W (conversely making a D5w stronger with a

D10). It just will not happen.

Now if you had any crystal or solid Dextrose (sugar) you could add

the correct amount to the D5W or the D10W to make a D12W, but that

would be DIFFERENT PROBLEM…….WOULDN'T IT??

So……………………that leaves diluting the D20W with either the D5W or the

D10W.

Use the Alligation method in my Tutorials.

I will not take the time to place the exact method here (plus it is

a difficult one to show on a post because the text and lines of the

Tic Tac Toe do not line up well on the posts. However the final

results:

20% - 5% = 15 parts

12 – 10 = 2 parts D20W

20 – 12 = 8 parts D10W

7/15 x 750 = 350ml and 8/15 x 750 = 400ml

Use 350 ml of D20W and 400 ml of D5W: Makes 750 ml of D12W

OR

20% - 10% = 10 total parts

12 - 5 = 7 parts of D20W

20 – 12 = 8 parts of D5W

2/10 x 750 = 150 ml and 8/10 x 750 ml = 600 ml

Use 150 ml of D20W and 600 ml of D10W: Makes 750 ml of D12W

The above took 10 minutes to place on the site.

OJT about 2 - 3 minutes.

All OJT times are mu best 'guestamates' and using paper pencil and

calculator.

I do hope that this has helped you as and educator and others who

are studying for the exam. If any of my answers do not 'jive' with

the answers in the Mathmasters book, please let me know. I did do

this quickly as I have to get back to some other tasks.

Alss let us say that these are what I would consider advanced

compounding calculations in my classroom. And I ahve no clue as to

whether or not any of this is on the PTCB exam. But if I did I could

only say that there is similar type problems and not give the exact

ones.

Respectfully,

Jeanetta Mastron CPhT BSChem

Pharm Tech Educator

Founder/Owner

> >From " Math Master Pharmaceutical Calculations for the Allied

Health Professional " 2nd edition by Noah Reifman

>

> 1. A drug label states the addition of 8 ml of saline will result

in a drug concentration of 250mg/ ml. You realized that you added

13ml to the vial. What is the final concentration in the vial?

>

>

>

> 2. Pharmacy has 300 ml of a 50% sol. 200 ml is added to this

sol. to decrease the concentration. How many grams of active

ingredients would there be in 4 oz of this diluted sol?

>

>

>

> 3. The tech must weigh out 20 grams of menthol and is told to

dissolve it in sterile water to make a 5% solution. The RPH then

receives an order to change the concentration to a 1% sol. What

will be the final volume of the 1% solution.

>

>

> 4. You have D5W, D10W, and D20W in stock. What quantity of any

two solutions should be used to make 750 ml of a D12W solution?

>

>

> Thanks for your help,

> Janet

>

>

>