2 pharmacy calculation questions

[Caverject]

I can't seem to figure this crap out...someone please enlighten me

thanks


A 0.94 kg infant is to be started on Prostin VR at a rate of 0.05 mcg/kg/min to run at 0.75 mL/h. Prostin VR comes in a concentration of 500 mcg/ml. How many mL's of the Prostin will be necessary to make 30 ml of the solution?


A patient received 15 mL of 10% KCl injection in one liter of 0.9% NaCL over 24 hours. Approximately how many mEq of chloride did the patient receive?
(atomic weights: Na = 23; K=39, Cl = 35.5)

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[PharmDstudent]

Is the answer to the first question ________?

 

[WVUPharm2007]

A 0.94 kg infant is to be started on Prostin VR at a rate of 0.05 mcg/kg/min to run at 0.75 mL/h. Prostin VR comes in a concentration of 500 mcg/ml. How many mL's of the Prostin will be necessary to make 30 ml of the solution?

Ok, one at a time....

0.94kg (.05mcg/kg/min) = 0.047mcg/min

0.75ml/h = 0.0125ml/min

So you'd need a concentration of 0.047mcg/0.0125mL

You need 30mL. So 0.0125ml:30L is 1:2400 ratio. Ergo, 0.047mg(2400) = 112.8 mcg of Prostin / 30mL solution.

Ergo, 112.8/500 per mL = 0.2256mL of the 500mcg/mL solution of Prostin.

 

[WVUPharm2007]

A patient received 15 mL of 10% KCl injection in one liter of 0.9% NaCL over 24 hours. Approximately how many mEq of chloride did the patient receive?
(atomic weights: Na = 23; K=39, Cl = 35.5)

10% = 10g/100ml. Ergo, 15ml/100ml (10g) = 1500mg of KCL

.9% NACL = 9g/1000mL, Ergo = 9000mg NaCl

Now we have to figure out how many mEqs of each salt we have.

MW of NaCl = 58.5
MW of KCl = 74.5

So, in moles, we have:

1500 KCl / 74.5 = 20.13 mmol KCl
9000 NaCl / 58.5 = 153.85mmol NaCl

Because the valence of both ions is one, as an example, it would be both 20.13 mEq of Cl- and K+ from 20.13 mmoles of KCL.

So, just add the two together, 20.13 + 153.85 = 174-ish.

 

[Express08]

Actually, for question #1 - you wouldn't dilute it like that at all in actual practice.

If you were familiar with Prostin VR, it leaches plasticizers from containers (including syringes) when in direct contact.

You also can't measure that small of a quantity.

But, if this is a test question - you can only measure out to four decimals if you do serial dilutions - too much contact with plastic & too much "loss" in hubs, etc...

If you check the Prostin VR info, it gives 2 methods of dilution depending on your pump capability - if you can infuse at 0.75ml/hr - it describes diluting the 500mcg in 25-100ml of D5 or NS & calculating the infusion rate.

Otherwise, you use an infusion chamber & mix directly in that & infuse @ 2-4ml/hr (going below 2ml/hr is less accurate in pumps that have discrete infusion rates.

But - you might get points for arithmetic. No points for practicality....

The school loses points for a dumb question!

 

[WVUPharm2007]

Actually, for question #1 - you wouldn't dilute it like that at all in actual practice.

If you were familiar with Prostin VR, it leaches plasticizers from containers (including syringes) when in direct contact.

You also can't measure that small of a quantity.

But, if this is a test question - you can only measure out to four decimals if you do serial dilutions - too much contact with plastic & too much "loss" in hubs, etc...

If you check the Prostin VR info, it gives 2 methods of dilution depending on your pump capability - if you can infuse at 0.75ml/hr - it describes diluting the 500mcg in 25-100ml of D5 or NS & calculating the infusion rate.

Otherwise, you use an infusion chamber & mix directly in that & infuse @ 2-4ml/hr (going below 2ml/hr is less accurate in pumps that have discrete infusion rates.

But - you might get points for arithmetic. No points for practicality....

The school loses points for a dumb question!

Man, this is an academic test question. They haven't been in the real world for 20 years. Who cares about reality. We've got idealistic standards to uphold, dammit!

 

[PharmDstudent]

Good job WVU.


I'm so distant from this kind of stuff right now. Why do they teach this stuff in the first semester? Wow!

It's scary to think that people put off studying calculations until the very end before taking the NAPLEX...

 

[Express08]

Good job WVU.


I'm so distant from this kind of stuff right now. Why do they teach this stuff in the first semester? Wow!

It's scary to think that people put off studying calculations until the very end before taking the NAPLEX...

NAPLEX won't ask you something like #1 - as WVU said - too academic & would be completely wrong clinically. They'd rather have you be able to know how to dilute & set the rate both ways - so learn that. Arithimetic is a good thing to know though!

But, there might be some value in #2 in real life. Sometimes, if you're the one doing the adjusting of the tpn, you need to know why you're causing a hyperchloremic acidosis.

An easier way would be to look at the Chem-12 panel & ABGs though....... But, if you don't know the how & why those values came about, then learning them is a good thing.

 

[oncpharmd]

I can't seem to figure this crap out...someone please enlighten me

thanks


A 0.94 kg infant is to be started on Prostin VR at a rate of 0.05 mcg/kg/min to run at 0.75 mL/h. Prostin VR comes in a concentration of 500 mcg/ml. How many mL's of the Prostin will be necessary to make 30 ml of the solution?

This question makes no sense whatsoever. That's why you're having trouble. We need to know either: 1) the standard concentration that should be dispensed (see the post by express) or 2) the time frame that the 30 mL should last.

 

[oncpharmd]

10% = 10g/100ml. Ergo, 15ml/100ml (10g) = 1500mg of KCL

.9% NACL = 9g/1000mL, Ergo = 9000mg NaCl

Now we have to figure out how many mEqs of each salt we have.

MW of NaCl = 58.5
MW of KCl = 74.5

So, in moles, we have:

1500 KCl / 74.5 = 20.13 mEq KCl
9000 NaCl / 58.5 = 153.85mEq

NEXT, we need to find which percentage of this is just chloride.

So, in KCL, 35.5/74.5 = 47.65% Cl by weight.
In NaCl, 35.5/58.5 - 60.68% Cl by weight.

Thus,
20.13 mEq KCL (.4765 is Cl) = 9.59 mEq Cl
153.85 mEq NaCl (.6068 is Cl) = 93.36 mEq Cl

Add it up and 102.95 mEq of Chloride ions.

Wrong!

mEq is a measure of valence (charge). Therefore, 1 mEq of Cl- is equal to 1 mEq of KCl. Thus, the answer is 20 mEq (from KCl) plus 154 mEq (from 1 L NSS), which is 164 mEq.

On a side note, 1 L bag of NSS (0.9% NaCl) is 154 mEq. Remember that. It will serve you well.

 

[Caverject]

Wrong!

mEq is a measure of valence (charge). Therefore, 1 mEq of Cl- is equal to 1 mEq of KCl. Thus, the answer is 20 mEq (from KCl) plus 154 mEq (from 1 L NSS), which is 164 mEq.

On a side note, 1 L bag of NSS (0.9% NaCl) is 154 mEq. Remember that. It will serve you well.

The answers available are 20, 36, 92,154, and 174.

Trust me, the questions are from a relevant source

Mike- thanks for the hook up in question 1

 

[MountainPharmD]

thumbdownthumbdownthumbdown

 

[WVUPharm2007]

Wrong!

mEq is a measure of valence (charge). Therefore, 1 mEq of Cl- is equal to 1 mEq of KCl. Thus, the answer is 20 mEq (from KCl) plus 154 mEq (from 1 L NSS), which is 164 mEq.

Yeah, you're right up until the point where you can't add.

I accidentally put mEqs of KCL when I figured it up when what I was doing was figuring mmols...so that threw off the rest of the problem. And I treated mEqs like mmols so I tried to find the ratio of ions. If it was an ion like iron, you'd have to take that step, but because the charges are (1), you can just use the raw salt mEqs.

So you'd have 153.85 + 20.13 = 174mEqs. That's my bad. I never take my damned time.

Oh well. I'll rewrite it above...either way, you think 154 + 20 is 164...heh....like a grown man being tackled by a fly...

 

[oncpharmd]

The answers available are 20, 36, 92,154, and 174.

Trust me, the questions are from a relevant source

Mike- thanks for the hook up in question 1

Ooops, typo on my part. 154 + 20 = 174

 

[SpirivaSunrise]

On a side note, 1 L bag of NSS (0.9% NaCl) is 154 mEq. Remember that. It will serve you well.

Just understanding the basics like that...you can get a big chunk of questions on the naplex without even working out the long arithmetic. If the patient's already getting a liter of saline (that's 154 mEq right there), there's only one answer left to account for the chloride.

Same with osmolarity...remember that body fluids contain 280-300 mOsm. If they ask you to calculate something that's isotonic, it'll be in that range. 1/2NS (hypotonic) will be around 150 mOsm, etc.

 

[jpharm445]

Wrong!

mEq is a measure of valence (charge). Therefore, 1 mEq of Cl- is equal to 1 mEq of KCl. Thus, the answer is 20 mEq (from KCl) plus 154 mEq (from 1 L NSS), which is 164 mEq.

On a side note, 1 L bag of NSS (0.9% NaCl) is 154 mEq. Remember that. It will serve you well.

mEq = milliequivalents. I think you're thinking of meV = millielectronvolts, which is actually a unit of energy. Your statement is correct in terms of milliequivalents since 1 mEq KCl yields 1 mEq Cl- (as opposed to 1 mEq of CaCl2, which yields 2 mEq Cl- for example). If you're talking charge, though, that statement makes no sense to me.

 

[oncpharmd]

mEq = milliequivalents. I think you're thinking of meV = millielectronvolts, which is actually a unit of energy. Your statement is correct in terms of milliequivalents since 1 mEq KCl yields 1 mEq Cl- (as opposed to 1 mEq of CaCl2, which yields 2 mEq Cl- for example). If you're talking charge, though, that statement makes no sense to me.

*Sigh*

Either you have an electrical engineering/physics background that you're trying to apply here or you're trying to sound smart.

To quote my Pharmaceutical Calculations textbook, the milliequivalent is a "unit of measure [related] to the total number of ionic charges in solution, and it takes note of the valence of the ions. In other words, it is a unit of measurement of the chemical activity of an electrolyte."

So you see, I'm talking charge in terms of chemical activity, not electrical activity. If this does not make sense to you, I suggest taking CHEM 101 over again.

 

[jpharm445]

*Sigh*

Either you have an electrical engineering/physics background that you're trying to apply here or you're trying to sound smart.

To quote my Pharmaceutical Calculations textbook, the milliequivalent is a "unit of measure [related] to the total number of ionic charges in solution, and it takes note of the valence of the ions. In other words, it is a unit of measurement of the chemical activity of an electrolyte."

So you see, I'm talking charge in terms of chemical activity, not electrical activity. If this does not make sense to you, I suggest taking CHEM 101 over again.

Your calculations book explains mEq correctly in saying that it is RELATED to the total number of charges and TAKES NOTE OF the valence, but it is not a measure of those properties. If you had said that in your reply, I wouldn't have said anything. The term "equivalent" can also apply to nonionic substances. For example, hydrolysis of acetic anhydride (1 equivalent) yields 2 equivalents of acetic acid. I could also say that dehydration of acetic acid (1 equivalent) yields 0.5 equivalents of acetic anhydride. Equivalents is a relative term; it's not a real measure of anything.

Why do you say that I'm trying to sound smart? Because I corrected your response? You corrected WVU by starting with "Wrong!" How is that a better approach? You're being very defensive and taking a simple correction far too personally. It is not necessary. Please relax.

 

[jpharm445]

This question makes no sense whatsoever. That's why you're having trouble. We need to know either: 1) the standard concentration that should be dispensed (see the post by express) or 2) the time frame that the 30 mL should last.

You're also incorrect here on both counts. 1) The concentration to be dispensed can be determined with the given information. WVU calculates it correctly. See his post. 2) The time to dispense 30 mL can be easily calculated from the given rate of 0.75mL/h. 30mL/(.75mL/h) = 40 h

Forgive me if I'm trying to show off my background in basic math and common sense.

 

[FruitFly]

wrrrrrrrrrrrrong! (I say that all the time when I volunteer tutored Elementary School kids. They're probably all in High School by now and obnoxious as hell thanks to me)

 

[nnguyenc]

what is the answer to number 2?

i still got a different answer than everyone else