Math problem------HELP ME PLEASE

[John Miller]

[h=3]Hi to all,

 

Here is the problem;

 

A pharmacist has calculated that a patientrequires 40 mmol of phosphate and 90 mEq of potassium to be added to the TPN.How many milliliters of potassium phosphate and how many milliliters ofpotassium chloride will be required?

 

 

 

Can you please help me solve this problem?

[/h]

[John Miller]

Hi to all forum member.

 

Please help me solve this problem.

 

Thanks,

[Prav05]

Hi John,

 

This is how I would solve. Please provide us the answer.

 

KCl-----has 2meq/ml (1meq of K + 1meq of Cl)

 

2meq---1ml

90 meq---?

 

90/2 ===45ml

 

K2PO4 ----has 3meq(2meq of K + 1meq of Po4)

 

MEQ === mmol x valency

meq== 40 x2

meq==80meq

 

1ml-----3meq

?-------80meq

 

80/3 === 26.7ml

 

so, kcl ===45ml & K2PO4 ==26.7ml

 

correct me.

[Prav05]

slight correction

K2PO4-----(2Meq K + 1Meq PO4)

MEQ== mmol x Valency

Valency of K is 1

 

MEQ = 40mmol x1

= 40 meq

 

K2PO4 ===3meq/ml

 

1ml-----3meq

?---------40meq

 

ml === 40/3 ===13.3ml of K2PO4

[B Kassa]

How we are addressing the fact that we are getting K both from k phosphate and K chloride? Are we not end up giving high dose of K by calculating the about of K without considering the contribution from each?

 

 

My understanding is equivalent weight is molecular wt divided by either the positive or negative electrical charges hence for K2PO4, the equivalents would not be moles multiplied by two 2 rather than 3 (K2PO4 ===3meq/ml)?

 

 

John Miller, could you provide us with the answer so that we could see the possibilities.

[spg2005]

Yes we have to consider that we get K from K phosphate and K chloride.

So first calculate the ml of K phosphate required that will supply 40 mMol of phosphate.

Normally the K phosphate that is supplied for TPN dilution are concentrated and in each ml we get ' 3 mMol of phosphate and 4.4 mEq of K '. (You can search this in google K phosphate for TPN http://prodruginfo.com/formulary/Forms/BSR-MR-TPN-001%201-07.pdf)

Also the K chloride for inj used in TPN preparations, Each ml contains 2mEq of K chloride USP.

I think we need to memorize the above information and use it for solving this problem.

Ans 13.3 ml of k phophate will supply 40 mMol of phosphate.

we know that 1ml of k Phophate will supply 4.4 mEq of K because that is how it is supplied by the manufacturer.

Therefore 13.3 ml should give us 58.52 mEq of K from K phosphate.

The remaining phosphate we can get from K chloride which is 31.48 mEq.

We get 2mEq of k chloride from each ml, therefore we need 15.74ml of K chloride to supply 31.48 mEq of K.

In conclusion we need 13.3ml of K phosphate and 15.74ml of K chloride.

Please correct me if I am wrong. There is a similar question in rxprep book.

[B Kassa]

That is an excellent explanation. It is very important to get the correct amount of K as excess level of K is very dangerous. Thanks everyone.

[Prav05]

It totally makes sense.....Thank you so much spg2005.