can someone work this math prob out for me?

[hookemhorns]

If boric acid has an E value of 0.52, what is the % strenght of an isotonic Boric Acid solution?

Answer: 1.73% (only I don't know how to come up with that)

Thanks for your help in advance....

---

Members don't see this ad.

 

[vicky336]

So this is my understanding of isotonicity - correct me if I'm wrong:

An E value means 1g of Boric Acid = 0.52g of NaCl

Since our standard isontonic solution is NS, we'll always figure out how many grams of NaCl are in that solution and work from there. This equation doesn't give us a particular volume so I'll assume 100mL which means there are 0.9g of NaCl in NS to make it isotonic. Based on my E value, I have to figure out how much boric acid I need to get enough "stuff" to equal 0.9g of NaCl.

1g boric acid/0.52g NaCl = x boric acid/0.9g NaCl

x=1.73g

Since I had assumed 100mL of solution, that gives me my answer of 1.73%. Hope this helps.

Another good equation to keep in mind if they don't give you E is:

E=(MW of NaCl)(dissociation factor of drug)/(MW of drug)(1.8)

 

[slowly707]

So this is my understanding of isotonicity - correct me if I'm wrong:

An E value means 1g of Boric Acid = 0.52g of NaCl

Since our standard isontonic solution is NS, we'll always figure out how many grams of NaCl are in that solution and work from there. This equation doesn't give us a particular volume so I'll assume 100mL which means there are 0.9g of NaCl in NS to make it isotonic. Based on my E value, I have to figure out how much boric acid I need to get enough "stuff" to equal 0.9g of NaCl.

1g boric acid/0.52g NaCl = x boric acid/0.9g NaCl

x=1.73g

Since I had assumed 100mL of solution, that gives me my answer of 1.73%. Hope this helps.

Another good equation to keep in mind if they don't give you E is:

E=(MW of NaCl)(dissociation factor of drug)/(MW of drug)(1.8)

LOL...all you have to is 0.9%/.52 = 1.73

 

[ucrx]

LOL...all you have to is 0.9%/.52 = 1.73

Yes, but that doesn't mean the OP will know how to solve it the next time.

 

[hookemhorns]

thanks vicky336...I was actually just making a really silly mistake. I don't think I've looked at these calculations since P1 year lol. For the E value I was setting 1 g NaCl to 0.52 g Boric Acid. That won't happen on the Naplex thanks again

 

[ucrx]

thanks vicky336...I was actually just making a really silly mistake. I don't think I've looked at these calculations since P1 year lol. For the E value I was setting 1 g NaCl to 0.52 g Boric Acid. That won't happen on the Naplex thanks again

Its so easy to do things like that! Luckily if you do it once and realize it, you might not do it again (when it counts!)

 

[johnep34]

60 years since I did isotonics and osmosis. Gather there are published lists of equivalents to NS, will have a look in Wiki.
Was a recent case where 10 x sodium phosphate soln supplied to lung clinic. Pt received undiluted and died agonising death. Case progressing through courts.
http://news.bbc.co.uk/1/hi/england/west_midlands/8426645.stm
johnep

 

[pharmaceuticsguy]

Isotonic boric acid solution means a solution in which you have only water and boric acid, but the tonicic strength is equal to that of 0.9% of sodium chloride.

0.9% means each 100 ml of this solution contains 0.9 grams of sodium chloride. Now, we need to find out the weight of boric acid that is equivalent to 0.9g of NaCl. Just by dividing the gram weight of NaCl with E value of an alternative excipient we can determine the weight of that excipient required to make the solution tonicically equal to NaCl. In this example, therefore we are dividing 0.9g with 0.52, and the answer is 1.73 grams. Now, this many grams of boric acid should be present in each 100ml, therefore, the percentage strength is 1.73%.

Hope I could clarify everything.

I published a video today which may help you in this regard: